iiir'^ 


I 


SYKES 
C©MStOCK 


GIFT  OF 
Publisher 


EDUCATION  DEPT. 


PLANE    GEOMETRY 


PLANE   GEOMETRY 


By 
MABEL  SYKES 

Instructor  in  Mathematics,  Bowen  High  School.  Chicago 
Author  of  "A  Source  Book  of  Problems  for  Geometry" 


and 


CLARENCE   E.  COMSTOCK 

Professor      of      Mathematics, 
Bradley  Polytechnic  Institute 


RAND  M9NALLY  &   COMPANY 

CHICAGO  NEW  YORK 


3^ 


c^ 


Copyright,  1918,  by  {  ,♦  |  r 

Rand  M?Nally  St  Company 
Edition  of  1922 


Kf 


c-22 


THE  CONTENTS 

The  Preface ix 

Chapter  I.     Introductory  p^^g 

Points  and  Straight  Lines 1 

Circles 5 

Angles 6 

Summary  and  Supplementary  Exercises 15 

Chapter  II.     Congruent  Triangles 

Introductory  Definitions 20 

Tests  for  Equal  Angles  and  Equal  Segments 21 

Application  of  Congruent  Triangles  to  Constructions  ....  30 

Nature  of  Theorems  and  Proofs 35 

Miscellaneous  Theorems  and  Exercises 37 

Chapter  III.     Parallels,  Perpendiculars, 
Angles,  Angle-Sums 

Introductory 48 

Parallels  . 50 

Angles  Made  by  Parallels  and  Transversals 54 

Angles  in  Triangles 59 

Angles  in  Polygons 63 

Miscellaneous  Theorems 66 

Supplementary  Exercises 69 

Chapter  IV.     Quadrilaterals 

Symmetry 75 

Parallelograms 79 

Special  Quadrilaterals 84 

Parallels  and  Segments  on  Transversals 88 

Supplementary  Exercises 94 

Chapter  V.     Inequalities 

Assumptions  for  Combining  Inequalities 99 

Fundamental  Tests  of  Inequality 100 

Tests  for  Unequal  Sides  and  Angles  in  One  Triangle    ....  102 

Tests  for  Unequal  Sides  and  Angles  in  Two  Triangles        .      .      .  105 

Supplementary  Exercises 106 

V 


571779 


vi  THE   CONTENTS 

Chapter  VI.     Circles  and  Related  Lines 

Introductory 108 

Related  Arcs,  Chords,  and  Central  Angles 110 

Chords  in  General Ill 

Tangents 114 

Two  Circles  and  Related  Lines 117 

Supplementary  Exercises 120 

Chapter  VII.     Circles  and  Related  Angles 

Relation  between  Central  Angles  and  Their  Arcs 123 

Relation  between  Inscribed  Angles  and  Their  Arcs       .      .      .      .  126 
Relation  between  Angles  Formed  by  Tangents  and  Chords  and 

Their  Arcs 135 

Summary  and  vSupplementary  Exercises 137 

Chapter  VIII.     Loci 

General  Considerations 143 

Loci  of  Points 146 

Determination  of  Points  by  the  Intersection  of  Loci    ....  152 

Loci  of  Centers  of  Circles 153 

Supplementary  Exercises 155 

Chapter  IX.     Ratio  and  Proportion 

Measurement  of  Segments 161 

Ratios 163 

Theory  of  Proportion 165 

Ratios  of  Segments  Made  by  Parallels    ........  168 

Similar  Triangles : 176 

Important  Special  Cases 182 

Applications  of  Equal  Ratios 191 

Summary  and  Supplementary  Exercises 197 

Chapter  X.     Area  and  Equivalence 

Introductory 208 

Measurement  of  Polygons 211 

Equivalent  Polygons 220 

Summary  and  Supplementary  Exercises 228 

Chapter  XL     Similarity 

Introductory 240 

Tests  for  Similar  Polygons 240 

Properties  of  Similar  Polygons 244 

Summary  and  Supplementary  Exercises 248 


THE   CONTENTS  vii 

Chapter  XII.    Regular  Polygons 

Definition 252 

Construction  of  Regular  Polygons 252 

Properties  of  Regular  Polygons 259 

Similar  Regular  Polygons 262 

Summary  and  Supplementary  Exercises 264 

Chapter  XIII.     Measurement  of  the  Circle 

The  Circumference  of  the  Circle 269 

Areas  of  Circles,  Sectors,  and  Segments        275 

Ratios  and  Circles 276 

Summary  and  Supplementary  Exercises 277 

Chapter  XIV.     Maxima  and  Minima 

Introductory 284 

Triangles 284 

Polygons  in  General 287 

Regular  Polygons 290 

Notes  on  Arithmetic  and  Algebra 292 

Tables 298 

Outline  Summary 301 

Index 309 


THE  PREFACE 

This  book  is  written  with  the  firm  conviction  that  it  is 
possible  to  give  to  high-school  young  people  a  more  sys- 
tematic training  in  the  science  of  geometry  than  is  furnished 
by  any  textbook  on  the  market  to-day.  In  this  connection 
the  two  main  features  of  the  book  should  be  noted : 

1.  The  analytical  method  of  attack  is  employed  throughout. 

Analyses  of  proofs  serve  several  purposes.  When  it  has 
been  found  by  actual  classroom  experience  that  any  particu- 
lar proof  is  so  difficult  that  the  pupil  cannot  reasonably 
be  expected  to  think  it  out  for  himself,  the  analysis  gives 
him  at  the  outset  the  gist  of  the  argument,  calls  his  attention 
to  the  method  of  proof  employed,  and  gives  him  some  idea 
of  how  the  proof  may  have  been  originally  invented.  In 
such  cases  not  only  is  the  analysis  given,  but  as  much  of 
the  proof  as  experience  has  found  necessary.  In  this  con- 
nection the  treatment  of  the  proofs  of  Theorems  3,  4,  9,  14, 
64,  77,  113,  and  120  may  be  noted  and  compared  with  the 
usual  treatment  of  these  same  proofs. 

More  important,  however,  is  the  fact  that  the  pupil  can, 
with  proper  training,  invent  many  of  his  own  proofs.  While 
nothing  can  do  away  entirely  with  the  element  of  inspiration 
in  getting  originals,  the  fact  remains  that  the  method  of 
analysis  is  the  method  by  which  every  trained  mind  attacks 
difficulties.  The  teacher  should  see  to  it  that  the  pupil  is 
continually  asking  himself  the  necessary  questions,  that  he 
sees  clearly  how  each  step  follows  from  the  preceding,  and 
that  the  results  are  set  down  in  the  orderly  form  here  em- 
ployed. The  statements  given  in  the  analyses  may  seem 
formal,  but  clearness  and  definiteness  are  essential.  No 
statement  should  be  permitted  which  does  not  clearly  indi- 
cate that  the  pupil  sees  all  the  essential  steps  of  the  argu- 
ment. The  proof  should  in  every  case  be  obtained  by 
working  backward  from  the  analysis. 


X  THE  PREFACE 

2.  'J  he  work  is  so  arranged  as  to  throw  emphasis  on  the 
important  theorems  and  methods. 

Without  emphasis  effective  analysis  is  impossible.  It  is 
said  to  be  a  fundamental  characteristic  of  the  mind  that 
any  lasting  impression  of  a  vast  field  requires  distinctions  in 
emphasis.  Moreover,  it  is  just  here  that  much  of  our 
geometry  teaching  has  failed.  Attention  is  called  to  the 
following  points  in  arrangement  and  presentation: 

a)  The  division  into  chapters  is  based  on  the  important 
general  ideas  in  geometry,  such  as  congruence,  loci,  ratio, 
area,  equivalence,  and  similarity.  If  the  work  is  so  pre- 
sented that  each  chapter  is  made  to  serve  the  special,  definite 
purpose  intended,  many  of  the  details  in  both  analysis  and 
proof  for  later  theorems  may  be  left  to  the  pupil.  Other- 
wise such  details  should  be  given.  The  purpose  of  chapter  ii, 
for  example,  is  to  train  pupils  in  the  use  of  congruent  tri- 
angles. If  this  purpose  has  been  accomplished,  the  pupil 
can  work  out  for  himself  the  details  for  such  congruent 
triangle  work  as  that  used  in  the  proofs  of  Theorems  33,  36, 
and  37.  Similarly,  it  is  only  on  the  assumption  that  chapter 
vii  and  §§209-212  have  served  their  purpose  that  the 
analysis  and  proof  for  Theorems  103,  104,  and  105  may 
be  safely  left  to  the  pupil. 

b)  The  purpose  of  each  chapter  is  carefully  worked  out 
in  the  order  and  grouping  of  the  theorems  and  exercises. 
In  the  minds  of  the  pupils  the  importance  of  a  theorem 
depends  solely  upon  the  frequency  with  which  it  is  used. 
To  this  end  the  dependence  of  the  minor  theorems  upon 
the  more  fundamental  ones  is  made  evident,  and  all  exercises 
given  in  connection  with  the  various  theorems  are  intended 
to  illustrate  the  use  of  those  special  theorems.  On  pages 
50-54,  for  example,  Theorems  10,  11,  and  12  depend  directly 
on  Theorem  9,  while  §67  consists  entirely  of  exercises  in 
which  it  is  required  to  prove  two  lines  parallel.  The  pur- 
pose of  chapter  ix  is  to  train  pupils  in  the  use  of  equal  ratios. 


THE  PREFACE      >  xi 

The  outline  of  this  chapter  and  the  arrangement  and  group- 
ing of  the  exercises  in  it  should  be  noted. 

c)  Theorems  and  problems  whose  interest  is  largely  theo- 
retical and  historical  or  which  have  no  important  place  in 
the  plan  of  the  work  are  inserted  in  the  supplementary  exer- 
cises at  the  end  of  the  chapters.  These  are  marked  with  a 
dagger  (f)  and  can  be  given  when  it  is  desired  to  extend  the 
course  or  to  prepare  for  the  examinations  of  the  College 
Entrance  Examination  Board. 

Inasmuch  as  the  traditional  order  is  the  most  convenient, 
it  has  been  preserved  except  where  emphasis  required  a 
change*.  This  accounts  for  the  separation  of  the  work  on 
ratio,  proportion,  and  similarity.  This  subject  involves 
two  points.  Pupils  should  not  only  be  familiar  with  the 
tests  for  similar  figures  and  the  properties  of  similar  figures, 
but  should  be  able  to  prove  ratios  equal.  The  work  is 
therefore  divided  into  two  chapters.  The  chapter  on 
similarity  is  given  after  the  chapter  on  area  and  equivalence 
to  permit  of  grouping  together  all  theorems  involving 
properties  of  similar  figures. 

Attention  is  also  called  to  certain  minor  features: 

The  introduction  is  natural  and  interesting,  concise  and 
to  the  point.  The  nature  of  the  exercises  in  this  first 
chapter  should  be  noted,  as  well  as  the  preliminary  use  of 
paper  folding  in  construction  work. 

The  algebraic  form  of  statement  for  theorems  and  exer- 
cises is  extensively  used.  See  especially  chapters  ix,  x, 
and   xi. 

The  formal  theory  of  limits  is  omitted.  The  idea  of  a 
limit  is  presented  informally  by  exercises,  but  proofs  that 
are  either  lacking  in  rigor  or  are  too  difficult  for  the  pupil 
are  omitted  entirely.  The  treatment  of  the  measurement 
of  the  circle  will  be  found  satisfactory  and  comprehensive. 

There  is  a  great  variety  of  exercises  with  concrete  setting. 
These  include  exercises  taken  from  surveying,  physics, 
architecture,    and   industrial   design.     For   illustrations   of 


xii  THE  PREFACE 

exercises  in  surveying,  see  pages  27,  46,  72,  191,  and  219; 
for  exercises  from  physics,  see  page  106;  for  exercises  from 
architecture,  see  pages  157,  158,  204,  206,  265,  and  280; 
for  exercises  from  industrial  design,  see  pages  69,  71,  97,  230, 
231,  and  235.  Illustrations  of  similar  problems  will  be 
found  on  pages  141,  142,  214,  266,  and  279. 

The  Notes  on  Arithmetic  and  Algebra  and  the  Outline 
Summary  preceding  the  Index  will  be  found  convenient 
for  reference.  Attention  is  also  called  to  the  unusually 
complete  character  of  the  index. 

Thanks  are  due  Professor  G.  A.  Miller  of  the  University 
of  Illinois  for  his  criticisms  of  the  historical  notes. 

M.    S. 

C.    E.    C. 
Chicago,  Illinois 
May,  iQi8 


PLANE    GEOMETRY 

CHAPTER  I 
Introductory 

1.  Geometry  treats  of  points,  lines,  surfaces,  and  solids. 
Plane  geometry  deals  with  lines  and  points  on  a  plane 

surface.  A  plane  surface  is  a  flat  surface,  like  a  plain.  In 
fact,  years  ago  the  name  was  "plain  geometry,"  that  is,  the 
geometry  of  the  plain.  In  the  early  part  of  the  seventeenth 
century  the  spelling  was  changed  from  "plain  geometry"  to 
"plane  geometry."  The  word  geometry  comes  from  two 
Greek  words  meaning  * '  the  earth ' '  and  ' '  to  measure. ' '  From 
earliest  times  the  results  of  geometry  have  been  used  for 
practical  purposes,  such  as  building  and  surveying.  The 
work  of  the  early  Egyptians  furnishes  an  illustration. 

POINTS  AND   STRAIGHT  LINES 
CONCRETE   REPRESENTATION 

2.  We  represent  a  point  on  paper  by  a  dot  made  with  a 
sharp-pointed  pencil  and  designate  it  by  a  capital  letter 
placed  near  the  dot  (see  point  A,  Fig.  1).  ^^ 

We  represent  a  straight  line  on  paper  by  a  ** 

mark  made  with  a  sharp-pointed  pencil  and  a         ^^^-  ^ 
straight  ruler  and  designate  it  by  a  small  letter  placed  on 
the  mark  (see  line  n,  Fig.  1). 

A  straight  line  resembles  a  tightly  stretched  string,  such 
as  a  thread  by  which  a  weight  is  suspended.  The  word 
line  comes  from  the  Latin  word  linea  meaning  "a  linen 
thread."  The  word  straight  comes  from  the  Anglo-Saxon 
verb  meaning  "to  stretch." 


'jE ;  :  ; :_ , ".  p.L^m  GEOMETRY 

LOCATION  OF  STRAIGHT  LINES 

3.  Ex.  1.  How  many  straight  lines  can  be  drawn  through  (1) 
one  given  point?  (2)  any  two  given  points?  (3)  any  three  given 
points?     (4)  any  four  given  points?    Illustrate  answers  by  figures. 

If  two  points  are  given,  the  sljraight  line  passing  through 
these  points  is  said  to  be  located  definitely. 
When  two  points  are  given,  as  points  A  and      ^  ^ 

B  (Fig.  2),  it  is  often  best  to  designate  the  Fig.  2 

line  passing  through  them  as  the  line  AB,  rather  than  by  a 
small  letter.  This  method  of  designating  the  line  locates  it 
definitely  with  respect  to  the  two  given  points  A  and  B. 

We  shall  assume  as  apparent  that 

Only  one  straight  line  can  pass  through  two  given  points. 

Ex.  2.  Designate  each  of  the  points  in  the  figures  that  you 
drew  for  Ex.  1  by  a  capital  letter.  Read  each  of  the  lines  by 
naming  two  of  its  points. 

LOCATION  OF  POINTS 

4,  Ex.  1.     How  many  points  are  there  in  a  straight  line? 

Ex.  2.  What  is  the  greatest  number  of  points  in  which  two 
straight  lines  can  intersect? 

Ex.  3.     Draw  two  straight  lines  with  no  possible  intersections. 

Ex.  4.  What  is  the  greatest  number  of  points  in  which  three 
straight  lines  can  intersect? 

Ex.  5.  Draw  three  straight  lines  with  (1)  no  possible  inter- 
sections; (2)  only  one  possible  intersection;  (3)  only  two  possible 
intersections;  (4)  three  intersections. 

Ex.  6.  Name  the  greatest  number  of  points  in  which  four 
straight  lines  can  intersect.     Draw  a  figiire  to  illustrate  your  answer. 

If  two  given  straight  lines  intersect,  the  point  of  inter- 
section is  said  to  be  located  definitely.  When  two  inter- 
secting straight  lines  are  given,  such  as  lines  a  and  6,  it  is 
often  best  to  design 2'*:e  the  point  of  intersection  as  the  point 


INTRODUCTORY  3 

ab  rather  than  by  one  capital  letter.  In  Fig.  3  the  two 
straight  lines  a  and  b  were  first  given;  their  intersection 
then  locates  point  0,  which  may  be 
called  the  point   ab.     This   method  of  ^^  o 

designating  the  point  locates  it  definitely    ^^^^         ^""^ 
with  respect  to  the  two  given  Hnes.  Fig.  3 

We  shall  assume  as  apparent  that 
Two  different  straight  lines  can  intersect  in  only  one 
point. 

Ex.  7.  Designate  each  of  the  lines  in  the  figures  you  drew  for 
Exs.  5  and  6  by  a  small  letter.  Read  each  of  the  points  whose  loca- 
tion is  determined  by  these  lines;  show  how  each  point  is  located. 


STRAIGHT-LINE   SEGMENTS 

6.  The  i)ortion  of  a  straight  line  terminated  by  two  given 
points  of  the  line  is  called  a  straight-line  segment.  Here- 
after the  single  word  segment  will  be  used  to  indicate  a 
straight-line  segment. 

If  two  segments  are  parts  of  the  same  straight  line,  they 
are  said  to  be  collinear. 

6.  We  transfer  segments   from  one  Hne  to  another  by 
the  use  of  the  compasses, 
the  dividers,  or  a  strip  of 
paper. 

To  transfer  a  given 
segment.  Open  the  di- 
viders to  the  segment 
required  by  laying  them 
on  the  given  segment. 
Without  changing  the  adjustment  place  one  leg  on  point 
A  and  mark  off  the  segment  AB  on  the  line  c   (Fig.  4). 

Ex.  1.     Find  a  segment  which  is  the  sum  of  two  given  seg- 
ments. 


4  PLANE  GEOMETRY 

Ex.  2.  Find  a  segment  which  is  the  difference  between  two 
given  segments.  Can  any  segment  be  subtracted  from  any  other 
segment? 

Ex.  3.    Show  how  to  multiply  a  given  segment  by  3 ;  by  5 ;  by  w. 

Note.  The  problem  of  dividing  a  segment  by  any  given  number 
is  not  as  simple  as  the  three  foregoing  problems.  It  will  be  studied 
later.     In  §9  we  learn  how  to  divide  a  given  segment  by  2. 

7.  Two  segments  are  said  to  be  congruent  if  they  can 
be  placed  upon  each  other  so  as  to  fit  exactly.  To  make 
two  segments  coincide,  their  extremities  must  be  made  to 
coincide. 

We  shall  assume  as  apparent  that 

Only  one  segment  can  be  drawn  joining  two  given  points. 

If  the  end  points  of  a  segment  are  known,  the  segment 
is  located  definitely. 

8.  A  segment  has  a  definite  length.  The  length  of  a 
segment  may  be  obtained  by  the  successive  application 
of  some  standard  unit. 

The  length  of  a  segment  joining'two  given  points  is  often 
called  the  distance  between  the  two  given  points. 

Segments  that  have  the  same  length  are  called  equal 
segments.  Equal  segments  are  congruent  and  congruent 
segments  are  equal. 

9.  To  find  the  mid-point  of  a  given  segment  by  paper 
folding.      Fold    the    paper    so   that    point 

A    falls  on   point    B.      Crease   the  paper  ^ ^ 

sharply.     Hold    it   to   the    light    and    see  Fig.  5 

that  one  part  of  the  segment  falls  exactly 
upon  the  other   part.     The   crease  marks  the  mid-point 
of  the  segment. 

We  shall  assume  as  apparent  that 

A  given  segment  has  only  one  mid-point. 

Exercise.    Divide  a  given  segment  into  four  eqval  parts. 

1 


INTRODUCTORY  5 

RAYS 

10.  A  portion  of  a  line  which  starts  at  a  given  point  and 
extends  indefinitely  in  a  given  direction  is 

called  a  ray.    The  point  is  called  the  origin  ^  ° 

of  the  ray  (see  ray  a,  origin  O,  Fig.  6).  ^^^-  ^ 

Two  rays  that  have  a  common  origin  and  extend  in  the 
same  direction  are  said  to  be  coincident.  If  they  have  the 
same  origin  and  extend  in  opposite  directions,  they  are 
collinear. 

A  number  of  rays  from  the  same  origin  form  a  pencil  of 
rays.     Make  a  drawing  showing  a  pencil  of  rays. 

11.  If  the  origin  and  any  other  point  of  a  ray  are  known, 
the  ray  is  located  definitely. 

We  shall  assume  as  apparent  that 
Only  one  ray  can  be  drawn  having  a  given  origin  and 
passing  through  a  second  given  point. 

CIRCLES 

12.  A  closed  curved  Une  every  point  of  which  is  equally 
distant  from  a  given  point  in  the  same  plane  is  called  a  circle. 
The  given  point  is  called  the  center  of  the  circle.  In  Fig.  7, 
O  is  the  center  of  the  circle. 

A  segment  drawn  from  the  center  to  the 
circle  is  called  a  radius.  A  segment 
drawn  through  the  center  and  terminating 
in  the  circle  is  called  a  diameter.  In  Fig. 
7,  OC,  OEy  and  OD  are  radii  and  DE  is 
a  diameter  of  circle  0,  ^^^-  '^ 

A  segment  joining  two  points  on  a  circle  is  called  a  chord 
of  the  circle.     In  Fig.  7,  XY  and  DE  are  chords  of  circle  0. 

A  part  of  a  circle  is  called  an  arc.  In  Fig.  7  the  part  of 
the  circle  between  points  A  and  B  is  an  arc  of  circle  0. 

Note.  Arc  comes  from  the  Latin,  and  means  "a  bow";  chord, 
from  the  Greek,  and  means  "the  string  of  a  musical  instrument." 


6  PLANE  GEOMETRY 

If  two  circles  can  be  made  to  coincide,  they  are  said  to  be 
congruent. 

Circles  with  equal  radii  are  congruent. 

The  truth  of  this  fact  may  be  shown  thus:  Draw  two 
different  circles  with  equal  radii.  Cut  them  out.  Put  the 
same  pin  through  the  center  of  each  circle. 

We  shall  assume  as  apparent  also  that 
Congruent  circles  have  equal  radii. 

Note.  A  circle  is  usually  drawn  with  the  compasses.  In  place 
of  compasses  a  pencil  and  a  thread  or  a  piece  of  chalk  and  a  string 
can  be  made  to  answer  the  purpose. 

ANGLES 

DEFINITIONS 

13.  A  figure  formed  by  two  rays  which  have  the  same 
origin  is  called  an  angle.  The  rays  are  called  the  sides  or 
arms  of  the  angle.  The  origin  of  the  rays  is  the  vertex  of 
the  angle. 

An  angle  may  be  designated  in  various  ways. 


Fig.  8 

In  Fig.  8  we  have  the  following  angles,  reading  from  left 
to  right:  LA,  Zab,  /.a,  Z2,  ABAC.  Notice  that  in  the 
last  case  the  letter  at  the  vertex  of  the  angle  is  read  between 
the  other  two  letters. 

An  angle  may  be  considered  as 
formed  by  the  rotation  of  a  ray  about 
its  origin.  The  size  of  the  angle  de- 
pends upon  the  amount  of  rotation. 
In  Fig.  9  which  is  the  larger  angle?     Why? 


INTRODUCTORY  7 

ADJACENT   ANGLES 

14.  Angles  that  have  a  common  vertex  and  a  common 
side  which  separates  the  angles  are  called  adjacent  angles. 
Make  a  drawing  to  illustrate  this  definition. 

Ex.  1.  How  many  angles  are  formed  when  a  ray  starts  from 
a  point  in  a  given  straight  line?  Illustrate  your  answer  by  a 
drawing  and  designate  the  angles  in  as  many  ways  as  possible. 

Ex.  2.  How  many  angles  are  formed  when  two  straight  lines 
intersect?  Illustrate  your  answer  by  a  drawing  and  designate 
the  angles  in  as  many  ways  as  possible. 

Ex.  3.  In  the  drawings  that  you  made  for  Exs.  1  and  2,  how 
many  pairs  of  adjacent  angles  can  you  find?  Can  you  find  angles 
that  are  not  adjacent? 

CONGRUENT   ANGLES 

16.  Two  angles  are  said  to  be  congruent  if  they  can  be  so 
placed  that  their  sides  are  coincident. 

To  construct  an  angle  that  shall  be  congruent  to  a  given 
angle. 

Method  I.  By  means  of  tracing  paper.  The  details  of 
this  method  are  left  to  the  pupil. 

Method  II.     By  means  of  an  angle-carrier.     Fasten  two 
pieces  of  cardboard  or  very  stiff  paper  together 
as  shown  in  Fig.   10.     Open  the  instrument 
until   the  edges   a   and   b  coincide  with  the 
sides  of  the  given  angle.     Without  changing 
the  adjustment  of  the  angle-carrier  transfer       i^^io.  lu 
the   instrument   to    the   desired    position    and    draw    the 
angle  formed. 

16.  Two  angles  may  be  added  by  placing 
them  adjacent  to  each  other.  The  angle 
formed  by  the  two  exterior  arms  is  the  sum  of 
the  two  adjacent  angles.  In  Fig.  11,  Z3  is 
the  sum  of  Z  1  and  Z  2.      Z  3  -  Z  1  +  Z  ?.. 


8 


PLANE   GEOMETRY 


A  smaller  angle  may  be  subtracted  from  a  larger  angle 

by  placing  the  smaller  inside  the  larger  so  that  they  have 
a  common  vertex  and  a  common  side.  The 
remaining  part  of  the  larger  angle  is  the 
difference  between  the  two  angles.  In  Fig. 
12,  Z3  is  the  difference  between  Zl  and  Z2. 
Z3=Z1-Z2.  Fig.  12 

Ex.  1.  Draw  two  angles  that  are  not  congruent  and  construct 
an  angle  equal  to  their  sum. 

Ex.  2.  Draw  two  angles  that  are  not  congruent  and  construct 
the  angle  equal  to  the  difference  between  the  larger  and  the  smaller 
angle. 

Ex.  3.  Construct  an  angle  that  is  twice  as  large  as  a  given 
angle. 

Ex.  4.  At  point  0  on  line  AB  draw  a  ray 
that  shall  make  with  the  line  AB  an  angle 
congruent  to  Z 1 .  Can  this  line  have  more 
than  one  position?     (Fig.  13.) 


o 
Fig.  i; 


17.  To    construct    the    bisector   of    a  ^ 
given  angle. 

To  construct  the  bisector  of  ZBAC. 

By  paper  folding.  Fold  the  paper  on 
a  line  extending  through  the  vertex  A 
(Fig.  14)  so  that  the  ray  AB  will  be 
coincident  with  the  ray  AC.  Crease 
the  paper  sharply.  The  crease  bisects 
ZA.     Why? 

We  shall  assume  as  apparent  that 

Only  one  ray  can  be  drawn  bisecting  a  given  angle. 

Ex.  1.     Divide  a  given  angle  into  four  congruent  parts. 

Ex.  2.  Draw  two  angles  that  are  not  congruent.  Construct 
an  angle  that  is  one-half  the  sum  of  the  two  angles  just  drawn. 

Ex.  3.  Draw  two  angles  that  are  not  congruent.  Construct 
an  angle  that  is  one-half  the  difference  obtained  by  subtracting 
the  smaller  angle  from  the  larger. 


Fig.   14 


INTRODUCTORY  9 

RIGHT  ANGLES   AND   PERPENDICULARS 

18.  When  a  ray  starts  from  a  point  in  a  straight  line  and 
forms  two  congruent  angles,  the  angles  are  called  right 
angles,  and  the  ray  is  said  to  be  perpen- 
dicular to  the  line.  In  Fig.  15,  Z 1  is 
congruent  to  Z  2,  Z 1  and  Z  2  are  called 
right  angles,  and  BC  is  said  to  be  per- 
pendicular to  Z)  A .  ^''''  ^^ 

In  Fig.  15  the  ray  BA  may  be  supposed  to  rotate  about 
point  B  as  origin.  One  complete  rotation  would  carry  it 
back  to  the  position  BA.  A  right  angle  is  obtained  by  one- 
quarter  of  a  complete  rotation. 

The  ray  OA  (Fig.  16)  may  start  from 
the  position  OA  and  rotate  until  it  ex- 
tends in  a  direction  exactly  opposite  to 
its  original  position.  It  has  made  one- 
half  of  a  complete  rotation.  The  figure 
AOAf,  is  called  a  straight  angle.  The  angle  formed  by  one 
complete  rotation  is  called  a  perigon. 

An  angle  less  than  a  right  angle  is  called  an  acute  angle. 

An  angle  greater  than  a  right  angle  and  less  than  a  straight 
angle  is  called  an  obtuse  angle. 

19.  To  construct  a  perpendicular  to  a  line  from  a  given 
point  in  the  line. 

To  construct  a  perpendicular  to  AB  from  point  O. 

By  paper  folding.  Fold  the  paper  through  point  0  so 
that  the  ray  OA  is  coincident  with  the  ray  OB.  Crease 
the  paper  sharply.    The  crease  is  _L  AB.    Why?  (Fig.  17.) 


Fig.   17 
Note.     In  ordinary  practice  a  perpendicular  to  a  line  from  a  point 
in  the  line  is  usually  drawn  by  means  of  a  rectangular  card  or  a  drafts- 
man's triangle.     Place  one  edge  of  the  card  on  line  ^45  with  the  corner 
of  the  card  at  point  O  and  draw  a  line  along  the  other  edge  of  the  card. 


10  PLANE  GEOMETRY 

We  shall  assume  as  apparent  that 

Only  one  perpendicular  can  be  drawn  to  a  line  from  a 
point  in  the  line. 

20.  To  construct  a  perpendicular  to  a  line  from  a  point 
not  in  the  line. 

To  construct  a  perpendicular  to  line  /  from  point  O. 

By  paper  folding.     Fold  the  paper  through  point  O  (Fig. 
18)   so  that  line  /  will  fall  upon  itself.  .^ 

Crease  the  paper  sharply.     The  crease  is 

the  perpendicular  to  line  /  from  point        j 

O.      Why?  Fig.   18 

Ex.  1.  Show  how  to  construct  a  perpendicular  to  a  line 
from  a  point  not  in  the  line  by  means  of  a  rectangular  card  or  a 
draftsman's  triangle. 

We  shall  assume  as  apparent  that 

Only  one  perpendicular  can  be  drawn  to  a  line  from  a 
point  not  in  the  line. 

Ex.  2,  How  many  ±s  can  be  drawn  to  a  given  line?  How 
may  one  of  these  be  located  definitely? 

21.  To  construct  the  perpendicular  bisector  of  a  given 
segment. 

To  construct  the  perpendicular  bisector  of  AB, 

By  paper  folding.     Fold  the  paper  so  that  point  A  falls 

on    point    B.     Crease    the     paper 

sharply.     The    crease    is    the    per-       j ~ B 

pendicular  bisector  of  AB.     Why?  fig.  19 

(Fig.  19.) 

22.  We  shall  assume  as  evident  that : 
I.  All  straight  angles  are  congruent. 

II.  All  right  angles  are  congruent. 
Ex.  1.     Show  how  to  bisect  a  straight  angle. 
Ex.  2.     Construct  by  paper  folding  an  angle  that  is  one-fourth 
of  a  straight  angle. 


INTRODUCTORY  11 

MEASUREMENT    OF   ANGLES 

23.  An  angle  is  measured  by  the  successive  application  to 
it  of  some  other  angle  considered  as  a  unit. 

Sometimes  angles  are  measured  by  comparing  them  with 
a  right  angle:  Thus  an  angle  may  be  H  of  a  right  angle; 
H  oi  Si  right  angle;  ^  of  a  right  angle.  For  most  purposes, 
however,  the  right  angle  is  too  large  a  unit. 

A  good  practical  unit  for  measuring  angles  is  -g-J^  of  a 
perigon  and  is  called  a  degree  (°). 

Ex.   1.     How  many  degrees  in  a  straight  angle?     in  a  right 
angle? 

Ex.  2.     How  many  degrees  in /^,  H,  Ks,  ^3'^4,  ^^  of  a  perigon? 

Ex.  3.     How  many  degrees  in  }^i,   %,  H,  H,  H  oi  a.  right 


angle 


Each  degree  may  be  divided  into  60  congruent  parts 
called  minutes  (')•  Each  minute  may  be  divided  into 
60  congruent  parts  called  seconds  ("). 

The  number  that  tells  how  many  times  the  unit  angle  is 
contained  in  a  given  angle  is  the  measure  or  the  measure 
number  of  the  given  angle. 

Angles  that  have  the  same  measure  number  are  said  to 
be  equal.  Since  congruent  angles  can  be  so  placed  that 
their  sides  take  the  same  direction,  congruent  angles  repre- 
sent the  same  amount  of  rotation  and  have  the  same  measure 
number.  Also  angles  that  have  the  same  measure  number 
represent  the  same  amount  of  rotation  and  can  be  made  to 
coincide.  In  other  words,  equal  angles  are  congruent  and 
congruent  angles  are  equal. 

Ex.  4.  The  sum  of  three  angles  is  360°.  The  second  is  three 
times  the  first,  and  the  third  is  four  times  the  first.  Find  the 
number  of  degrees  in  each.  Can  you  construct  a  figure  by  paper 
folding  to  illustrate  your  answer? 


12  PLANE  GEOMETRY 

IMPORTANT   SPECIALLY  RELATED   ANGLES 
24.  Two  angles  are  called  complementary  if  their  sum  is 
an  angle  of  90°.     Each  of  two  complementary  angles  is 
called  the  complement  of  the  other. 

Ex.  1.     Find  the  complements  of  the  following  angles: 

a.  60°  d.  62°  27'  g,  33° 

b.  44°  e.  59°  18'  h.  42°  2' 

c.  39°  10'  /.  25°  20'  i.  ic° 

Note:  Many  of  the  exercises  that  follow  can  and  should  be  solved 
by  an  algebraic  equation. 

Ex.  2.  If  the  complement  of  an  angle  is  K  of  the  angle,  find 
the  number  of  degrees  in  the  angle  and  its  complement. 

Ex.  3.  If  an  angle  is  H  of  its  complement,  find  the  number 
of  degrees  in  the  angle  and  its  complement. 

Ex.  4.  Draw  any  acute  angle.  Construct  the  angle  which  is 
the  complement  of  the  first  angle. 

Ex.  5.  Draw  two  equal  acute  angles.  Draw  the  complement 
of  each  of  these  angles.  Cut  out  these  complements  and  place 
one  upon  the  other.    Are  they  congruent? 

Ex.  6.  How  are  the  complements  in  Ex.  5  obtained  by  sub- 
tracting equal  angles  from  equal  angles? 

The  fact  illustrated  in  Exs.  5  and  6  will  be  assumed  as 
evident.     It  may  be  stated  as  follows: 

Complements  of  equal  angles  are  equal. 

26.  Two  angles  are  said  to  be  supplementary  if  their  sum 
is  an  angle  of  180°.  Each  of  two  supplementary  angles  is 
called  the  supplement  of  the  other. 

Ex.  1.     Find  the  supplement  of  each  of  the  following  angles: 

a.  75°  d.  59°  22'  g.  90°  21' 

b.  18°  25'  e.  63°  18'  h.  16°  18' 


c. 

11°  65' 

/.  12°  30' 

i.  x° 

Ex. 

2. 

How  many  ( 

degrees  in  an  angle  that 

is  H  of  its  supple- 

ment? 

Ex. 

3. 

An 

angle  is 

;  H  of  its  supplement. 

Find  the  number 

of  degrees  in  the  angle  and  its  supplement. 


INTRODUCTORY 


13 


Ex.  4.  Draw  any  angle.  Draw  the  angle  which  is  the  supple- 
ment of  the  first  angle. 

Ex.  5.  Draw  two  equal  angles.  Draw  the  supplements  of 
each  of  these  angles.  Cut  out  these  supplements  and  place  one 
upon  the  other.    Are  they  congruent? 

Ex.  6.  How  are  the  supplements  in  Ex.  5  obtained  by  sub- 
tracting equal  angles  from  equal  angles? 

The  fact  illustrated  in  Exs.  5  and  6  will  be  assumed  as 
evident.     It  may  be  stated  as  follows: 
Supplements  of  equal  angles  are  equal. 

26.  Certain  important  facts  concerning  sums  of  angles 
are  illustrated  in  the  next  three  exercises.  It  is  evident 
that  the  sum  of  all  the  parts  of  an  angle  is  equal  to  the 
whole  angle.  Show  which  is  the  whole  angle  and  what  are 
its  parts  in  each  of  these  three  exercises. 


Ex.  1.  If  in  Fig.  20  the  ray  OB  starts  from 
point  0  in  the  line  CA,  how  many  degrees  are 
there  in  the  sum  oi  A\  and  2? 


Ex.  2.     In  Fig.  21,  if  AOB  is  a  straight  line, 
what  is  the  sum  of  Z1+Z2+Z3+Z4? 


Ex.  3.     In  Fig.  22  what  is    the  sum  of 
Z1+Z2+Z3+Z4+Z5? 


Fig.  22 

The  facts  illustrated  in  Exs.  1-3  will  be  assumed  as  evi- 
dent.    They  may  be  stated  as  follows: 

I.  If  a  ray  starts  from  a  point  in  a  straight  line,  the  sum 
of  the  two  adjacent  angles  formed  on  one  side  of  the 
line  is  180°,  or  a  straight  angle. 

Such  angles  are  called  supplementary  adjacent  angles. 


14 


PLANE   GEOMETRY 


11.  The  sum  of  the  adjacent  angles  on  one  side  of  a 
straight  line  formed  by  any  number  of  rays  having  a  com- 
mon origin  on  the  line  is  180°,  or  a  straight  angle. 

III.  The  sum  of  the  adjacent  angles  formed  by  a  number 
of  rays  from  the  same  origin  is  360°,  or  a  perigon. 

Ex.4.  If  Zl+/2+Z3  =  180°,  and  Z2  is  twice  Zl,and  Z3 
is  three  times   Z2,  find  the  number  of  degrees  in  each  angle. 

Ex.  5.  If  four  equal  angles  form  a  perigon,  how  many  degrees 
in  each? 

Ex.  6.  If  six  equal  angles  form  a  perigon,  how  many  degrees 
in  each? 

Ex.  7.  If  Zl+Z2+Z3+Z4  =  360°,  and  Z4  is  twice  Z3, 
Z3  twice  Z2,  and   Z  2  twice  Zl,  how  many  degrees  in  each? 

27.  Statement  I  in  the  preceding  section  indicates  a  very 
close  relation  between  supplementary  adjacent  angles  and  a 
straight  angle.  The  next  two  exercises  illustrate  another 
phase  of  this  relation. 


Ex.  1.  Copy  two  right  angles  from  cards  or 
draftsman's  triangles  (Fig.  23).  Place  them  so 
that  point  0  falls  on  point  0',  side  OA  along  side 
O'A',  and  OB  opposite  O'B'.  What  kind  of  an 
angle  is   A  BOB'} 

Ex.  2.  Draw  any  two  equal  angles,  A 1 
and  2  (Fig.  24).  Construct  Z3,  the  supple- 
ment of  Z2.  How  are  Zl  and  3  related? 
Why?  Cut  out  the  three  angles.  Place  Z 1 
in  the  position  occupied  by  Z2.  What  kind 
of  an  angle  is  /.B'OC} 


Fig    24 


The  fact  illustrated  in  Exs.  1  and  2  will  be  assumed  as 
evident.     It  may  be  stated  as  follows: 

If  two  supplementary  angles  are  adjacent,  their  exterior 
sides  are  collinear.  , 


INTRODUCTORY  15 

28.  Two  an^'lcs  arc  called  vertical  or  opposite  if  the  sides 
of  one  are  prolongations  of  the  sides  of  the  other.  If  two 
lines  intersect,  two  pairs  of  vertical  angles  are  formed .  Thus 
in  Fig.  25  lines  h  and  k  intersect,  forming  vertical  angles, 
Z  1  and  Z  3,  also  Z  2  and  Z  4. 

Ex.  1.  Draw  two  intersecting  lines  (Fig.  25). 
Show  that  A2  and  4  are  the  supplements  of  the 
same  angle.  Show  that  A 1  and  3  are  the  supple- 
ments of  the  same  angle. 

Ex.  2.  In  Fv^.  25  why  is  Zl=Z3and  Z2=  Z4? 
In  general,  why  are  vertical  angles  equal? 


Fig.  25 


The  fact  illustrated  in  Exs.  1  and  2  will  be  assurned  as 
evident.     It  may  be  stated  as  follows: 

If  two  straight  lines  intersect,  the  opposite  or  vertical 
angles  are  equal. 

SUMMARY  AND    SUPPLEMENTARY   EXERCISES 

SUMMARY   OF   GEOMETRICAL   ASSUMPTIONS 

29.  The  foregoing  definitions  and  exercises  justify  the 
following  assumptions: 

A.  Concerning  the  definite  location  of  points: 

As.  1.  Two  different  straight  lines  can  intersect  in 
only  one  point,  or  two  intersecting  straight  lines  locate  a 
point  (§4). 

As.  2.     A  segment  can  have  only  one  mid-point  (§9\ 

B.  Concerning  the  definite  location  of  segments,  rays, 
and  straight  lines: 

As.  3.  Only  one  segment  can  be  drawn  between  two 
points,  or  a  segment  is  located  definitely  if  its  extremities 
are  given  (§7). 

As.  4.  Only  one  ray  can  be  drawn  having  a  given 
origin  and  passing  through  a  second  given  point  (§  11). 

As.  5.  Only  one  ray  can  be  drawn  bisecting  a  given 
angle  (§17). 


16  PLANE   GEOMETRY 

As.  6.  Only  one  straight  line  can  pass  through  two 
given  points  (§3). 

As.  7.  Only  one  perpendicular  can  be  drawn  to  a 
line  fjcom  a  given  point  in  the  line  (§19). 

As.  8.  Only  one  perpendicular  can  be  drawTi  to  a 
line  from  a  given  point  not  in  the  line  (§20). 

C.  Concerning  circles: 

As.  9.     Circles  with  equal  radii  are  congruent   (§12). 
As.  10.     Congruent    circles  have  equal    radii   (§12). 

D.  Concerning  equal  angles: 

As.  11.     All  straight  angles  are  equal  (§22). 

As.  12.     All  right  angles  are  equal  (§22). 

As.  13.     Complements  of  equal  angles  are  equal  (§24). 

As.  14.     Supplements  of  equal  angles  are  equal  (§25). 

As.  15.     Vertical  angles  are  equal  (§28). 

E.  Concerning  angle-sums: 

As.  16.  If  a  ray  starts  from  a  point  in  a  straight 
line,  the  sum  of  the  two  adjacent  angles  formed  on  one  side 
of  the  line  is  180°,  or  a  straight  angle  (§26). 

As.  17.  The  sum  of  the  adjacent  angles  on  one  side 
of  a  straight  line  formed  by  any  number  of  rays  having  a 
common  origin  on  the  line  is  180°,  or  a  straight  angle  (§26). 

As.  18.  The  sum  of  the  adjacent  angles  formed  by 
a  number  of  rays  from  the  same  origin  is  360°,  or  a 
perigon   (§26). 

F.  Concerning  straight  angles: 

As.  19.  If  two  supplementary  angles  are  adjacent, 
their  exterior  sides  are  coUinear  (§27). 

GENERAL  ASSUMPTIONS 
30.  The  following  assumptions  are  also  true: 

As.  20.  If  equal  segments  (or  angles)  are  added  to 
equal  segments  (or  angles),  the  results  are  equal  segments 
(or  angles). 


INTRODUCTORY  17 

As.  21.  If  equal  segments  (or  angles)  are  subtracted 
from  equal  segments  (or  angles),  the  results  are  equal  seg- 
ments (or  angles). 

As.  22.  If  equal  segments  (or  angles)  are  multiplied 
by  the  same  number,  the  results  are  equal  segments  (or 
angles) . 

As.  23.  If  equal  segments  (or  angles)  are  divided  by 
the  same  number,  the  results  are  equal  segments  (or  angles) . 

As.  24.  Segments  (or  angles)  that  are  equal  to  the 
same  segment  (or  angle)  are  equal. 

As.  25.  Equal  segments  (or  angles)  may  be  substi- 
tuted for  equal  segments  (or  angles). 

Note.  Hereafter  when  these  assumptions  are  used  they  should  be 
quoted  in  the  form  in  which  they  apply;  for  example:  Equal  segments 
may  be  substituted  for  equal  segments,  or  equal  angles  may  be  sub- 
stituted for  equal  angles,  as  the  case  may  require. 

MISCELLANEOUS   EXERCISES 

31.  After  solving  each  of  the  following  exercises  state 
clearly  and  in  full  the  definitions  or  assumptions  which  it 
is  intended  to  illustrate. 

1.  One  of  two  supplementary  angles  is  40°  15'  larger  than 
the  other.     Find  the  angles. 

2.  Find  an  angle  whose  complement  is   }4  as  large  as  the 
angle  itself. 

3.  Show  that  if  two  lines  intersect  so  that  one  angle  is  a  right 
angle  all  the  angles  are  right  angles. 

4.  Find  an  angle  whose  complement  is  }4  of  its  supplement. 
Can  you  construct  this  angle  by  paper  folding? 

5.  What  angles  do  the  hands  of  a  clock  make  at  5  o'clock? 
at  10  o'clock? 

6.  If,  in  Fig.  26,Z2=  Z6,  show  that 
fl.  Z  3=Z  6  (/.  Z  1=Z  8 
^>.  Z  2=Z  7  d.  Z  4=Z  5 
c.  Z  1=Z  5            /.   Z  4=Z  8 


18  PLANE   GEOMETRY 

7.  Which  of  the  following  directions  locate  the  required  line 
or  rays  definitely?    Why? 

a.  Construct  a  perpendicular  to  line  AB. 

b.  Construct  a  perpendicular  to  line  AB  from  point  O  outside 

of  AB. 

c.  Connect  points  C  and  D. 

d.  Bisect  A  A. 

e.  Draw  a  line  through  point  A. 

J.  From  point  O  outside  of  line  A  B  draw  a  ray  cutting  line 

AB. 
g.  Draw  a  ray  which  shall  make  a  given  angle  with  a  given 

line. 
h.  Draw  a  perpendicular  bisector  to  a  given  segment. 
i.  Draw  a  segment  bisecting  a  given  segment. 

8.  Two  rays  start  from  point  0  on  the  same  side  of  line  AB. 
Of  the  three  angles  formed  the  second  is  3  times  the  first  and  the 
third  is  5  times  the  first.     Find  the  angles. 

9.  Find  an  angle  whose  complement  is  25°  smaller  than  the 
angle  itself. 

10.  Can  you  find  an  angle  whose  complement  is  %  of  its  sup- 
plement? 

11.  Draw   four  straight   lines   so  that  there  will   be:     a.   No 

possible  intersections;  h.  Only  one  possible  intersec- 
tion; c.  Only  three  possible  intersections;  d.  Four 
possible  intersections;  e.  Five  possible  intersections; 
/.  Six  possible  intersections. 

12.  Can  you  draw   four   straight   lines  so   that  there  can  be 
only  two  possible  intersections? 

13.  How  many  degrees  in  the  supplement  of   /.x  \i  its  comple- 
ment is  23°? 


14.  In  Fig.  27  show  that  Z  H-  Z44-Z5+ZG- 
2rt.  ^.  What  other  combinations  of  angles  in 
Fig.  27  will  be  equal  to  two  right  angles? 


Fig.  37 


INTRODUCTORY 


19 


15.  Draw  two  complementary  adjacent  angles  and  the  bisec- 
tor of  each.  How  many  degrees  in  the  angle  made  by  the 
bisectors?    Why? 

16.  Draw  two  supplementary  adjacent  angles  and  the  bisec- 
tor of  each.  How  many  degrees  in  the  angle  made  by  the 
bisectors?    Why? 

17.  Draw  AABC  (Fig.  28)  so  that    Z1=Z2. 
Show  that  Z  3  =  Z  4. 

18.  How  many  degrees  in  the  complement  of  Z6  if     a 

the  supplement  of  Z  ft  is  110°? 

Fig.  28 


32.  ABBREVIATIONS   AND    SYMBOLS   USED 


Fig. 


figure 


Ex exercise 

Exs exercises 

Def definition 

As assumption 

Ass assumptions 

Th theorem 

Ths theorems 

Cor corollary 

alt.  int alternate  interior 

comp complementary 

sup supplementary 

adj Adjacent 

§ section 

.* therefore 

° degree 

' minute 

" second 

cm centimeter 

mm millimeter 

_  (equal  to 

(is  equal  to 

^  (congruent  to 

(is  congruent  to 

c^  (similar  to 

I  is  similar  to 


> (greater  than 

(is  greater  than 
<^  (less  than 

(is  less  than 

Z ,  A angle,  angles 

rt.  Z right  angle 

rt.  A right  angles 

(perpendicular  to 

-^ -^us  perpendicular 

'     to 
II  (parallel  to 

(is  parallel  to 

A,  A triangle, triangles 

rt.  A right  triangle 

n ,  CS  .  .  .  /rectangle,   rec-     ' 
(     tangles 

D ,  [s] square,  squares 

EJ,  CsJ.  .  .  (parallelogram 
(parallelograms 
y — ^   /-g-\        (trapezoid,  trap- 
'         ■  ■    (     ezoids 

O,  (D circle,  circles 

.15 arc^B 

AB chord  ^5 

5 area 

per perimrjter 


CHAPTER   II 

Congruent  Triangles 
INTRODUCTORY  DEFINITIONS 

33.  Any  two  figures  that  can  be  made  to  coincide  are 
called  congruent  figures. 

In  congruent  figures  corresponding  sides  or  angles  are 
sides  or  angles  that  coincide  or  that  can  be  made  to 
coincide. 

We  shall  add  the  following  to  the  list  of  assumptions: 

As.  26.  Any  figure  can  be  moved  about  in  space  with- 
out changing  either  its  size  or  its  shape. 

As.  27.  Figures  congruent  to  the  same  figure  are  con- 
gruent to  each  other. 

34.  A  figure  formed  of  three  segments  joined  end  to  end 
consecutively  is  called  a  triangle.  Such  a  figure  has  three 
sides,  three  angles,  and  three  vertices.  Unless  it  is  other- 
wise stated,  a  triangle  should  be  drawn  with  its  sides  of 
different  lengths  (Fig.  29). 


A  triangle  that  has  at  least  two  sides  equal  is  called  an 
isosceles  triangle.  The  angle  included  by  the  equal  sides 
is  called  the  vertex  angle,  and  its  vertex,  the  vertex  of  the 
triangle.     The  third  side  is  called  the  base  of  the  triangle. 

A  triangle  with  all  its  sides  equal  is  called  an  equilateral 
triangle. 

20 


CONGRUENT  TRIANGLES 


21 


Fig.  3] 


TESTS  FOR  EQUAL  ANGLES   AND 
EQUAL   SEGMENTS 

TESTS   I  AND   II  FOR   CONGRUENT  TRIANGLES 

35.  Ex.  1.  Draw  a  triangle  with  two  sides  equal  to  the  seg- 
ments a   and    b    (Fig.    30). 

Draw  this  triangle  with  a  soft  « 

pencil  on  paper  that  is  not  j, 

too    hesLvy.     Compare    your  Fig,  30 

figure  with  your  neighbor's 

by  placing  one  paper  upon  the  other  and  holding  them  to  the  light. 

Ex.  2.  Draw  a  tri- 
angle with  two  sides  equal 
to  the  segments  a  and  b 
and  the  included  angle 
equal  to  ZC  (Fig.  31). 
Compare  your  figure  with 

your    neighbor's    as    ex- 

plained  in  Ex.  1. 

Ex.  3.  Draw  a  triangle 
with  two  angles  equal  to  A  A 
and  B  and  the  included  side 
equal  to  segment  c  (Fig.  32). 
Compare  your  triangle  with  ^^ 

your    neighbor's    by   placing  \ 

one  paper  upon  the  other  and 

holding  them  to  the  light.  Fig~~32 

Ex.  4.  Draw  any  triangle  and  letter  it  ABC.  Draw  another 
triangle  and  letter  it  XYZ,  but  make  two  sides  and  the  included 
angle  of  AXYZ  equal  to  two  sides  and  the  included  angle  of 
AABC.    Compare  the  two  triangles. 

c 
Note.     In  Ex.  4  and  in   all  exercises  requiring 
the   construction   of   figures   the  pupil   should   not 
only  make  the  drawing  but  should  tell  how  he  did  a- 
it.      Sentences   should    be    short  and  exact.      For 
example  (Fig.  33): 

(1)  Make  ZX=/.A.     (2)  Make  XY  =  AB.  Etc.    ^     Fig.  33 


22 


PLANE   GEOMETRY 


Theorkm  1.*  If  two  sides  and  the  included  angle  of  one 
triangle  are  equal  to  two  sides  and  the  included  angle  of 
another  triangle,  the  triangles  are  congruent  in  all  corre- 
sponding parts  and  are  called  congruent  triangles. 


Fig.  34 


Given    AABC   and 
AC  =  DF. 


ADEF,    ZA=ZD,    AB  =  DE,  and 


To  prove  AABC  and  ADEF  congruent  in  all  correspond 
ing  parts. 

Proof: 

STATEMENTS 

1.  Place  AABC  on  ADEF  so 
that  yl  Swill  iallonDE,A 
on  D,  B  on  E,  and  C  and 
F  on  the  same  side  of  DE, 

2.  Segment  AC  will  fall  along 
the  line  of  DF. 

3.  Point  C  will  fall  on  point  F. 

4.  BC  will  coincide  exactly 
with  EF. 


REASONS 

1.  This  can  be  done 
because  AB  =  DE. 


2.    ZA=  ZD. 


5.  AABC  ^  ADEF. 


Note:     An  angle  of  a  triangle  is 
two  adjacent  sides. 

♦The  formal  proof  of  Th.   1  may  be 
teacher  desires. 


3.  AC  =  DF. 

4.  If  the  extremities  of 
two  segments  coin- 
cide, the  segments  will 
coincide  exactly. 

5.  Two  triangles  that 
coincide  exactly  are 
congiTient. 

said  to  be  included  between  its 
jjostponed   imtil   such   time  as  the 


CONGRUENT  TRIANGLES 


23 


36.  Theorem  2.*  If  two  angles  and  the  included  side 
of  one  triangle  are  equal  to  two  angles  and  the  included 
side  of  another  triangle,  the  triangles  are  congruent  in  all 
corresponding  parts  and  are  called  congruent  triangles. 


Fig.  35 


Given    AABC  and   ADEF,   /.A=  Z.D,    An=  Z.E,   and 
AB  =  DE. 

To  prove  A  ABC  and  ADEF  congruent  in  all  correspond- 
ing parts. 

Proof: 

STATEMENTS 

1.  Place  A.4^Con  ADEF  so 
that  AB  falls  on  DE,  A 
on  D,  B  on  E,  and  C  and  F 
on  the  same  side  of  DE. 

2.  AC  will  fall  along  the  line 
oiDF. 

3.  BC  will  fall  along  the  line 
of£F. 

4.  C  \^all  fall  on  the  line  of  DF 
and  also  on  the  line  of  EF, 

5.  .*.  C  will   fall  on  point  F. 


REASONS 

1.  This     can     be     done 
because ? 


2.  Why? 

3.  Why? 

4.  Why? 


6.   .*.  AABC  ^  ADEF. 


5.  Two  lines  can  inter- 
sect at  only  one  point, 
f).  Why? 


Note.    A  side  of  a  triangle  is  said  to  be  included  between  the  two 
angles  adjacent  to  i1. 


♦The  formal  proof  of  Th.  2  may  be  postponed  until  such  time 
teacher  desires. 


the 


24 


PLANE  GEOMETRY 


*  Exercise.  Draw  any  triangle  and  letter  it  ABC.  Draw 
another  triangle  and  letter  it  XYZ,  but  make  two  angles  and  the 
included  side  of  AXYZ  equal  to  two  angles  and  the  included 
side  oi  AABC.  In  how  many  ways  can  you  construct  a  triangle 
congruent  to  a  given  triangle? 

37.  Ex.  1.     Construct   an   isosceles  triangle  with 
base  3  cm.  and  each  of  the  equal  sides  5  cm.  (Fig.  36) . 

Note.     Notice  the  wording  for  the  directions  for  this    -*^ ^s 

drawing.  Fig.  36 

I.  Draw  segment  AB=S  cm. 
II,  With  A  as  a,  center  and  5  cm.  as  a  radius  make  an  arc  above  AB, 

Etc. 
Ex.  2.    Construct  an  equilateral  triangle  with  each  side  4  cm. 

Theorem  3.    The  angles  opposite  the  equal  sides  of  an 
isosceles  triangle  are  equal. 


Fig.  37 

Given  the  isosceles  A  ABC,  AC  =  BC, 

To  prove  AA^  LB. 

Analysts  and  construction: 

I.  To  prove   ZA=  ZB,  prove  A  A  and  B  correspond- 
ing angles  of  congruent  triangles. 

II.  To  obtain  the   triangles,  bisect   ZC,  continue   the 
bisector  CO  to  meet  the  base  at  0. 

III.  .-.   to  prove  ZA=ZB,  prove  AAOCmABOC. 

IV.  To  prove  AAOC  ^  ABOC,  prove  two  sides  and  the 

included  angle 


CONGRUENT  TRIANGLES 


25 


Proof: 


STATEMENTS 

SEASONS 

.a.  AC  =  BC. 

I. 

a.    • 

Let  the  r»uDi^ 

h.    Z1=Z2. 

c.  CO  =  CO. 

d,  :.AACO^ABOC. 

h. 
c. 

d.   \ 

give  the  rea- 
sons for  each 
statement. 

.  ZA=ZB. 

II 

C  or  re  s  p  ending 
angles   of   congruent 
triangles. 

Cor.  An  equilateral  triangle  has  all  of  its  angles  equal; 
that  is,  it  is  equiangular. 

38.  The  method  used  in  Th.  3  is  very  important.  One 
way  to  prove  two  segments  or  two  angles  equal  is  to  find 
two  triangles  that  contain  these  segments  or  these  angles 
and  to  prove  these  triangles  congruent.  In  general  we  say 
that 

Corresponding  sides  of  congruent  triangles  are  equal. 
Corresponding  angles  of  congruent  triangles  are  equal. 

In  the  following  exercises  use  the  analysis  and  proof  of 
Th.  3  as  a  model. 

To  make  an  analysis,  it  is  necessary  to  ask  one's  self  a 
series  of  questions.  Start  with  what  is  to  be  proved,  and 
by  means  of  these  questions  unravel  the  proof  step  by  step. 
In  Th.  3,  for  example,  you  should  begin  by  asking  yourself, 
What  are  the  tests  for  equal  angles?  The  answer  to  this 
question  gives  the  first  step  in  the  analysis  and  suggests  the 
construction  given  in  step  II.  You  should  then  ask  your- 
self, What  are  the  tests  for  congruent  triangles?  The  answer 
to  this  question  gives  step  IV  of  the  analysis.  This  process 
should  be  continued  for  each  theorem  or  exercise  until  the 
entire  proof  is  clear.  The  proof  is  the  analysis  done  back- 
ward, with  all  authorities  stated  in  full. 


26 


PLANE   GEOMETRY 


Ex.  1.     Construct  Fig.  38.     Make  AD  and  BC  perpendicular 
to  AB.     Find  0,  the  mid-point  of  AB.    Make 
AD  =  BC.    Join   DO  and  CO.     Prove   DO  =  CO. 
How  might   AAOD  be  made  to   coincide  with 
ABOC? 


Ex.  2.  Construct  Fig.  39.  Make  BC  perpen- 
dicular to  AO.  Make  Z  1  =  Z2.  Prove  AB  =  AC 
and  BX  =  XC.  How  might  AAXB  be  made  to 
coincide  with  AAXC? 


Ex.  3.  Construct  Fig.  40.  Draw  XY,  the  per- 
pendicular bisector  of  segment  AB.  Join  F,  any 
point  in  XY,  to  A  and  to  5.     Prove  PA=PB. 

Ex.  4.     Construct  Fig.   41.     ^i5  is  any  seg- 
ment.    Make     Z1=Z2   and    Z3=  Z4.    Which 

A 

sides  can  be  proved  equal?  Prove  them  equal. 
How  might  A^^A'  be  made  to  coincide  with 
AABY? 


Fig.  40 


Ex.  5.     Construct  Fig.  42. 

angle.     Draw  BO  bisecting  Z  B. 

Connect  A,  any  point   in  OB, 
Prove  ^A  =  CA. 


Draw  ABC,  any 

Make  AB  =  BC. 

with   A    and   C. 


Ex.  6.  Construct  Fig.  43.  ABC  is  an  isosceles 
triangle.  0  is  the  mid-point  of  AB.  AX  =  BY. 
Prove  OX  =  OY. 


Ex.  7.  Construct  Fig.  44.  Draw  any  seg- 
ment AB.  Find  O,  its  mid-point.  Draw  rays 
h  and  k  so  that  Z  1  =  Z  2.  Through  O  draw 
any  line  that  will  intersect  h  and  k.  Call  the 
])oints  of  intersection  D  and  C  respectively. 
Prove  that  AD  =  BC  and  that  DO  =  CO.  How 
might  ABOC  be  made  to  coincide  with  AAOD  ? 


Fig.  42 


43 


Fig.  44 


CONGRUENT   TRIANGLES  27 

Ex.  8.     Show   that   congruent  triangles  may  be  used  to  find 
the  distance  across  a  pond  as  follows  (Fig.  45) : 

First  set  up  a  stake  at  any  convenient  point,  as 
0,  from  which  points  A  and  B  are  both  visible.     Set 
up  a  stake  at  Z)  in  a  straight  line  with  O  and  A  so 
that  OD  =  0A.     In  the  same  way  place  stake  C  so  qz 
that  CO  =  OB.     Measure  CD.  Fig.  46 

Ex.  9.     Explain  from  Fig.  46  how  to   find      3^^^5^-^ 

the  distance  across  a  stream.     Notice  that  the      ^Sa^g^rrSv 

required  distance  AB  is  perpendicular  to  a  line  \J 

along  the  bank  of  the  stream.  ^ 

Fig.  46 

Ex.   10.     Show  that  the  distance  across  a  stream  may  be 
measured  as  follows  (Fig.  47):    A    pole  is  set  c 

up  at  A  with  a  stick  fastened  at  the  top  so  that  ■^'''''^Pv 

Z  DC  A  may  be  altered  by  moving  the  stick  CD.       / 
A    person    stands    with    the    eye    at    C    and  ^7         T^^^^^^^^"^ 
moves  the  stick  CD  until  he  just  sees  point  B.  Fig.  47 

The  pole  is  then  turned  around.  Standing  with  eye  at  C  and 
looking  along  CD'  he  locates  point  B'.  He  then  measures  AB' . 
Could  you  do  this  using  the  visor  of  your  cap  rather  than  a 
pole  and  a  stick?  Is  it  necessary  to  turn  through  180°?  This 
device  is  said  to  be  an  old  one.* 

Note.  Theorems  1  and  2  were  probably  known  to  Thales.  Thales 
is  regarded  as  the  founder  of  one  of  the  earliest  Greek  schools  of  mathe- 
matics, about  600  B.C.  It  is  said  that  Th.  2  was  used  in  those  days  to 
determine  the  distance  of  a  ship  at  sea.  It  is  not  known  how  this 
was  done;  but  a  tower  or  a  cliff  might  have  been  used  as  the  base  of 
a  triangle  of  which  the  ship  formed  the  vertex.  The  base  angles  of 
this  triangle  could  be  found  by  observation.  Show  from  Ex.  10  how 
the  solution  to  the  problem  could  be  completed. 

Ex.  11.     In  Fig.  48,  AB  and  CD  are   two  b 

intersecting    lines.     AO  =  OB    and    CO  =  OD.  ^\''y;^:^C\ 

Prove  that   AC  =  BD.     Join  CB  and   AD  and  ^^^            ° 

prove  CB  =  AD.  ^^°- ^^ 

♦See  W.  E.  Stark,  "Measuring  Instruments  of  Long  Ago,"  School 
Science  and  Mathematics,  1910, 


28 


PLANE   GEOMETRY 


Ex.  12.  Construct  Fig.  49.  Draw  AB  any  segment, 
struct  h  and  k  perpendicular  to  ^j5  at  A  and  B 
respectively.  Find  0,  the  mid-point  oiAB.  Make 
Z 1  =  Z  2  and  extend  the  sides  imtil  they  inter- 
sect h  and  k.  Call  the  points  of  intersection  D 
and  C  respectively.  Prove  that  AD  =  BC  and 
DO  =  OC. 

Ex.  13.  Construct  Fig.  50.  ABC  is  an  isosceles 
triangle.  O  is  the  mid-point  oi  AB.  Zl=  Z2.  Prove 
AX  =  BY  and  OX=-OY. 


Con- 


TEST  III  FOR   CONGRUENT   TRIANGLES 

39.  Ex.  1 .  Construct  two  triangles  ABC  and  X YZ  so  that  AB  = 
XF,  BC=YZj  and  CA=ZX.  Draw  these  triangles  with  a  soft 
pencil  on  two  pieces  of  fairly  thin  paper.  Compare  them  by 
placing  one  paper  on  the  other  and  holding  the  papers  to  the 
light.  Before  these  triangles  can  be  proved  congruent  what  must 
you  know? 

Ex.  2.    Construct  a  triangle  having  its  sides  equal  to  three 
given  segments. 

Theorem  4.  If  three  sides  of  one  triangle  are  equal  to 
three  sides  of  another  triangle,  the  triangles  are  congruent. 


B  V 


Fig.  51 

Given    AABC    and    ADEF,    AB  =  DE,    AC  =  DF,   and 
CB=^FE. 


To  prove  AABC  m  ADEF. 


CONGRUENT  TRIANGLES 


29 


Analysis  and  construction: 

I.  To  prove  AABC  ^ADEF,  prove  AC  =  /.F. 

II.  "  "  ZC=ZF,  place  AABC  so  that  AB,  the 
longest  side  of  AABC,  coincides  with  DE,  the 
longest  side  of  ADEF,  and  point  C  is  opposite 
point  F.  Join  CF  and  prove  that  ZC^and  ZF 
are  made  by  adding  the  base  angles  of  two  isos- 
celes triangles. 


Proof: 

STATEMENTS 

REASONS 

I.  a.  EF  =  EC. 

I. 

a.  Given. 

b.  AEFC  is  isosceles. 

b.  A    triangle    that 
has    two    sides 
equal  is  isosceles. 

c.    Z2  =  Z4. 

c.  The  base  angles 
of  an  isosceles  A 
are  equal. 

II.  a.  DF  =  DC. 

II. 

a.  Why? 

b.  ADFC  is  isosceles. 

b.  Why? 

c.    Z1  =  Z3. 

c.  Why? 

III.    .'.ZC=ZF. 

III. 

Equal  angles  added 
to    equal    angles 
give  equal  angles. 

IV.  ADEF  ^  ADEC. 

IV. 

Why? 

V.  ADEF  m  AABC. 

V. 

Triangles  congruent 
to  the  same  triangle 
are     congruent     to 
each  other. 

Ex.  3.     Prove  Th.  4  by  proving  that  ZA  =  ZD;  with 
Z  C  and  Z  F  obtuse  angles. 

Ex.  4.     Construct  Fig.  52.     ABC  is  an  isosceles  tri-^ — ^    ^ 
angle.    O  is  the  mid-point  of  ^ B.     Prove  Z1=Z2.  Fig.  52 


30  PLANE  GEOMETRY 

APPLICATION  OF  CONGRUENT  TRIANGLES 

TO   CONSTRUCTIONS 

CONSTRUCTION   OF  ANGLES 

40.  Problem  1.    At  a  given  point  in  a  given  line  to  con- 
struct an  angle  equal  to  a  given  angle. 


B  X 

Fig.  53 
Given  line  /,  point  X  in  line  /,  and  Z  BAG. 
To  construct  at  point  X  in  line  /  an  angle  equal  to  Z  BAG, 
Analysis  and  directions: 
I.  In  order  to  construct  an  angle  WXO  equal  to  /.A, 
construct  two  congruent  triangles  that  shall  contain 
A  A  andX 
II.  With  A  as  a  center  and  any  radius  cut  the  sides  of  ZA 
at  D  and  E. 

III.  With  X  as  a  center  and  the  same  radius  draw  an  arc 

cutting  Hne  /  at  W. 

IV.  With  1^  as  a  center  and  DE  as  a  radius  cut  the  last 

arc  at  O. 
V.  Join  X  and  O. 
Proof: 

STATEMENTS  REASONS 

I.  a.  AE  =  XO. 

b.  AD  =  XW.  Let  the  pupil  give  all 

c.  DE  =  WO.  reasons  in  full. 

d.  .'.ADAE^AWXO. 
II.   .-.  ,CA==ZX. 

Exercise.  How  is  it  possible  to  construct  the  ZX  in  more 
than  one  position  on  the  line  I  and  still  have  it  fulfill  the  require- 
ments? 


CONGRUENT  TRIANGLES  31 

DIVISION   OF  ANGLES 
41.  Problem  2.    To  construct  the  bisector  of  a  given 


angle. 


Given  ZBAC. 

To  construct  the  bisector  of   ZBAC. 

Analysis  and  directions: 

I.  In  order  to  construct  the  bisector  of  Z.A,  construct 

a  line  through  point  A  so  that  Z  1  equals  Z  2. 
II.  In  order  to  construct  Zl  =  Z2,  construct  congruent 
triangles  that  contain   A  1  and  2. 

III.  With  i4  as  a  center  and  any  convenient  radius  draw 

an  arc  cutting  the  sides  of  /.A  Sit  E  and  D. 

IV.  With  £  as  a  center  and  any  convenient  radius  draw 

an  arc. 
V.  With  Z)  as  a  center  and  the  same  radius  cut  this  last 

arc  at  O. 
VI.  Join  A  and  0. 
Let  the  pupil  give  the  proof. 

Ex.  1.     Show  that  a  carpenter's  steel  square 
may  be  used  to  bisect  an  angle  as  follows:    Mark  o 
off  equal  distances  OA  and  OB  on  the  sides  of 
the  angle.     Place  the  square  as  shown  in  Fig.  55. 
Mark  point  D.    Join  D  and  O. 

Fig.  55 

Ex.  2.     Construct    with   ruler  and  compasses  the  figures  for 
Exs.  4  and  5,  §38. 


32  PLANE  GEOMETRY 

Ex.  3.  Draw  two  vertical  angles  and  bisect  each.  Show  by 
As.  19  that  the  two  bisectors  are  collinear. 

Ex.  4.    Construct  one-half  the  supplement  of  any  given  angle. 

Ex.  5.  Construct  a  triangle  congruent  to  a  given  triangle.  In 
how  many  ways  can  this  be  done? 

Ex.  6.  Draw  any  triangle  and  bisect  each  angle.  Bisect  each 
angle  of  an  equilateral  triangle. 

42.  Note.  Trisection  of  Angles.  By  the  method  given  in  §41  we 
can  bisect  any  angle  that  we  choose.  The  trisection  of  an  angle  is  a 
much  more  difficult  problem.  We  shall  find  a  little  later  how  to 
trisect  a  right  angle.  In  elementary  geometry  we  confine  ourselves 
to  the  circle  and  straight  line  and  use  no  instruments  except  the  com- 
passes and  the  straightedge.  A  straightedge  is  a  ruler  that  is  not 
graduated  to  any  scale.  It  has  been  shown  that  angles  generally 
cannot  be  trisected  by  the  use  of  these  instruments  only.  The  Greeks 
learned  How  to  trisect  any  angle,  and  since  their  time  many  ways  of 
trisecting  angles  have  been  found.  These  methods,  however,  have 
always  required  other  curves  than  the  circle  and  other  instruments 
than  the  compasses  and  straightedge. 

Draftsmen's  methods  for  trisecting  angles  are  approximations. 


A  number  of  different  instruments  for  trisecting  angles  have  been 
made.  Fig.  56  shows  one  such  instrument.  What  segments  are 
made  equal?  Show  that  Fig.  54  is  used  twice  in  Fig.  56.  Why  is  the 
angle  trisected?  How  could  an  instrument  for  bisecting  angles  be 
made?  Such  an  instrument  is  sometimes  used  by  carpenters  for 
cutting  and  fitting  moldings. 


CONGRUENT  TRIANGLES  33 

CONSTRUCTION   OF  PERPENDICULARS 

43.  Problem  3.    To  construct  a  perpendicular  to  a  line 
from  a  given  point  in  the  line. 

/ 

/ 


t- 


V 

\ 

\ 


\. 


Fig.  57 

Given  line  /  and  point  0  in  line  /. 

To  construct  a  perpendicular  to  line  /  from  point  0. 

Analysis  and  directions: 

I.  In  order   to   construct   OX    ±    /  at   0,   construct 
Z1=Z2. 

II.  In  order  to  construct   Zl=  Z2,  construct  two  con- 
gruent triangles. 

III.  With  0  as  center  and  any  convenient  radius  draw 

an  arc  cutting  line  /  at  Y  and  Z. 

IV.  With  Y  as  center  and  a  longer  radius  draw  an  arc. 
V.  With  Z  as  center  and  the  same  radius  cut  the  last 

arc  at  X. 
VI.  Join  X  and  0. 

Let  the  pupil  give  the  proof  (see  §18  for  the  proof  to  I). 

Ex.  1.  Construct  at  point  0  in  a  given  line  /  an  angle  of  45°; 
of  135°;  of  22>^°;  of  157K°. 

Ex.  2.    Divide  a  given  angle  into  four  equal  parts. 

Ex.  3.    Construct  Fig.  39  with  ruler  and  compasses. 

Ex.  4.  Construct  the  complement  of  any  acute  angle.  Can 
the  complement  have  more  than  one  position? 

Ex.  5.  Construct  two  complementary  adjacent  angles.  Con- 
struct the  bisector  of  each.  How  many  degrees  in  the  angle  made 
by  the  bisectors? 


34  PLANE   GEOMETRY 

44.  Problem  4.    To  construct  a  perpendicular  to  a  line 
from  a  point  not  in  the  line. 

q 

\ 
4\ 


r\^ 


Fig.  58 

Given  line  /  and  point  0  not  in  line  /. 

To  construct  a  perpendicular  to  /  from  0. 

Directions: 

I.  With  O  as  center  and  any  radius  draw  an  arc  cut- 
ting /  at  y  and  Z. 
11.  With  Y  as  center  and  a  radius   greater  than  half 
YZ  draw  an  arc  opposite  O. 

III.  With  Z  as  center  and  the  same  radius  cut  this  arc 

at  X. 

IV.  Join  0  and  X. 

Analysis: 
I.  To  prove  OX  ±  /,  prove  Z  1  =  Z  2. 
II.    "      "       Z1=Z2,    join    OY    and    OZ    and    prove 

AYOAUAZOA. 
III.  To  prove  /\YOA  ^  AZOA,  prove  Z3=  Z4. 
IV.     '*      "        Z3=Z4,    ..... 
Let  the  pupil  complete  the  analysis  and  give  the  proof. 

Note.     The  proof  for  Prob.  4  will  be  much  simpler  after  intersecting 
circles  are  studied.     The  reasons  for  the  directions  are  then  apparent. 

Ex.  1.     Draw  any  triangle  and  construct  the   perpendicular 
from  each  vertex  to  the  opposite  side. 

^  Ex.  2.     Make  the  drawing  called  for  in  Ex.  1  for  an  equilateral 
triangle  and  also  for  a  triangle  containing  an  obtuse  angle. 


CONGRUENT  TRIANGLES  35 

46.  Pkoblkm  T).    To  construct  the  perpendicular  bisector 
of  a  given  segment. 


01  B 

i 
I 

I 

Fig.  59 
Given  the  segment  AB. 

To  construct  the  perpendicular  l:)isector  of  AB. 
Directions:  ^ 

I.  With  i4  as  a  center  and  any  radius  greater  than  H  AB 

construct  arcs  above  and  below  AB. 
II.  With  B  as  a  center  and  the  same  radius  intersect  these 

arcs  at  C  and  D. 
III.  Join  C  and  D. 
Let  the  pupil  give  the  analysis  and  the  proof. 

Note.  Like  the  proof  for  Prob.  4,  the  proof  for  Prob.  5  is  simpler 
after  one  studies  intersecting  circles. 

Ex.  1.  In  the  solution  for  Prob.  5,  which  radii  must  be  made 
equal?  Which  radii  need  not  be  made  equal?  Draw  the  figure 
and  give  the  directions,  analysis,  and  proof  for  this  problem, 
varying  the  radii  as  much  as  possible. 

Ex.  2.     Divide  a  given  segment  into  four  equal  parts. 

NATURE  OF  THEOREMS  AND  PROOFS 

46.  The  statements  of  geometrical  facts  in  this  and  the 
preceding  chapter  have  received  three  different  narnes: 
assumptions,  theorems,  and  corollaries. 

A  theorem  is  a  statement  of  a  fact  that  is  to  be  proved 
true. 

An  assumption  is  a  statement  of  a  fact  that  is  taken 
as  true  without  proof.     Its  ^mth   is  taken   for  granted. 


36  PLANE  GEOMETRY 

Many  of   the  assumptions  in  chapter  i  were  obtained  by 
observation.     Many  were  shown  to  be  true  informally. 

The  proof  of  a  theorem  shows  that  if  certain  facts  are 
known  to  be  true  a  certain  other  fact  must  be  true.  The 
facts  that  are  known  or  given  constitute  the  hypothesis. 
The  fact  to  be  proved  is  called  the  conclusion. 

In  Theorem  3  the  hypothesis  is:  The  triangle  has  two 
equal  sides.  The  conclusion  is:  The  angles  opposite  these 
sides  are  equal. 

In  Theorem  4  the  hypothesis  is :  Three  sides  of  one  triangle 
are  equal  to  three  sides  of  another  triangle.  The  conclusion 
is:  The  triangles  are  congruent. 

The  hypothesis  and  conclusion  of  a  theorem  may  be 
found  by  studying  the  grammatical  construction  of  the 
statement  of  the  theorem.  If  the  sentence  contains  a  clause 
beginning  with  "if,"  this  clause  is  the  hypothesis.  If  there 
is  no  such  clause,  the  complete  subject  is  the  hypothesis  and 
the  complete  predicate  the  conclusion. 

In  each  case  the  proof  consists  in  showing  that  the  con- 
clusion must  follow  from  the  hypothesis.  The  analysis 
shows  how  the  proof  has  been  or  may  be  thought  out. 
The  proof  is  the  analysis  worked  backward  and  is  set 
down  in  what  is  called  the  synthetic  form.  The  synthetic 
method  is  the  opposite  of  the  analytical  method. 

The  special  method  used  for  the  proofs  of  Ths.  1  and  2 
is  called  proof  by  superposition.  This  word  is  derived 
from  two  Latin  words.     What  is  its  literal  translation? 

A  theorem  that  follows  easily  from  another  theorem  is 
called  a  corollary  of  that  theorem. 

Proofs  are  required  for  all  theorems  and  exercises  that 
follow  unless  some  specific  statement  is  made  to  the  con- 
trary.    In  all  proofs  a  warrant  must  be  given  for  each  step. 

Some  of  the  exercises  have  been  called  problems.  Any 
geometrical  problem  calls   for  the  construction  of  some 


CONGRUENT  TRIANGLES  37 

geometrical  figure  to  fulfill  certain  stated  requirements. 
Unless  it  is  otherwise  stated,  these  constructions  must  be 
performed  with  compasses  and  straightedge  only.  In  all 
problems  it  is  necessary  to  prove  that  the  completed  figure 
fulfills  the  stated  requirements. 

MISCELLANEOUS  THEOREMS  AND   EXERCISES 
47.  OUTLINE  REVIEW 

A.  We  have  the  following  methods  for  proving  two  segments 

or  two  angles  equal : 

I.  In  the  case  of  segments  look  for 

a.  Sums,  differences,  equal  multiples,  or  equal 

parts  of  equal  segments. 

b.  Sides  of  isosceles  or  of  equilateral  triangles. 

c.  Corresponding  sides  of  congruent  triangles. 

11.  In  the  case  of  angles  look  for 

a.  Sums,  differences,  equal  multiples,  or  equal 

parts  of  equal  angles. 

b.  Right  angles. 

c.  Supplements  of  equal  angles. 

d.  Complements  of  equal  angles. 

e.  Vertical  angles. 

/.    Corresponding  angles  of  congruent  triangles. 
g.  Base  angles  of  an  isosceles  triangle. 

B .  Either  segments  or  angles  may  be  proved  equal  by  prov- 

ing them  corresponding  parts  of  congruent  tri- 
angles. 

To  prove  the  triangles  congruent  compare 

I.  Two  sides  and  the  included  angle  of  one  with  two 
sides  and  the  included  angle  of  the  other. 

II.  Two  angles  and  the  included  side  of  one  with  two 
angles  and  the  included  side  of  the  other. 

III.  Three  sides  of  one  with  three  sides  of  the  other. 


38  PLANE   GEOMETRY 

48.  Theorem  5.  If  a  perpendicular  be  erected  to  a 
straight  line,  oblique  segments  drawn  from  the  same  point 
in  the  perpendicular  cutting  the  straight  line  at  equal  dis- 
tances from  the  foot  of  the  perpendicular  are  equal. 


Hypothesis:    Line  XY l.iQ  line  I  and  the  oblique  segments 
CA  and  CB  are  drawn  from  Cso  that  AO  =  OB. 
Conclusion:    CA=CB. 
Analysis: 
I.  To  prove  CA=CB,  prove  AAOC  ^  ABOC. 

IL     "      **       AAOC  ^  ABOC,  compare 

The  proof  is  left  to  the  pupil. 

Exercise.  How  could  you  make  the  congruent  triangles  in 
Fig.  60  coincide? 

49.  In  the  following  exercises  segments  and  angles  are 
to  be  proved  equal.  The  analysis  given  for  Th.  3  may  now 
be  shortened  as  shown  under  Th.  5. 

Note.  The  figures  for  all  exercises  and  theorems  should  be  con- 
structed according  to  the  hypothesis  with  ruler  and  compasses. 

Ex.  1.  Construct  an  isosceles  triangle  ABC.  Let  CO,  the 
bisector  of  Z  C,  meet  the  base  A  B  at  0.  Prove  that  the  seg- 
ments joining  0  with  the  mid-points  of  ylC  and  BC  are  equal. 

Ex.  2.  Two  isosceles  triangles  stand  on  opposite  sides  of  the 
same  base.  Prove  that  the  segment  joining  the  vertices  bisects 
both  vertex  angles. 

Ex.  -3.  Investigate  the  case,  Ex.  2,  in  which  the  two  isosceles 
triangles  stand  on  the  same  side  of  the  same  base. 


CONGRUENT  TRIANGLES 


39 


50.  Theorem  6.     The  bisector  of  the  vertex  angle  of  an 
isosceles  triangle  is  the  perpendicular  bisector  of  the  base. 


AADC    is    isosceles,    AC  =  BC,    and     CO 
CO  is  a  ±  bisector  of  AB. 


Hypothesis: 
bisects    Z  C. 
Conclusion: 
Analysis: 
I.  To  prove  CO  a  _L  bisector  oi  AB,  prove 


II. 


prove 


Z2=  Z3. 
^AO==BO. 
Z2=  Z3 
AO  =  BO 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 

61.  Theorem  7.  The  segment  which  joins  the  vertex 
of  an  isosceles  triangle  with  the  mid-point  of  the  base 
bisects  the  vertex  angle  and  is  perpendicular  to  the  base 
(Fig.  61). 

Hypothesis:     A  ABC  is  isosceles,  AC  =  BC,  and  CO  joins 
the  vertex  C  with  0,  the  mid-point  of  base  AB. 
f  1.  CO  bisects  ZC. 
I  2.  CO±AB. 

Analysis: 

I.  To  prove  that  CO  bisects  ZC,  prove  Z5=  Z6. 
11.    '•        "        '•    CO±AB,  prove  Z2=  Z3. 

"^-  lz2=Z3    I'P^^^^ 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 


C  Otic  lus  ion 


40 


PLANE  GEOMETRY 


Fig.  62 


EXERCISES  INVOLVING   CONGRUENT   TRIANGLES 

62.  In  the  exercises  that  follow,  special  care  is  needed 
in  making  the  analysis.  Study  carefully  the  analyses  given 
for  Ths.  6  and  7  and  for  Ex.  1  below.  As  compared  with 
the  analyses  for  Ths.  3  and  5,  one  or  more  extra  steps  are 
required.  The  pupil  should  carefully  ask  himself  the 
proper  questions  at  each  step  and  should  be  perfectly  sure 
that  he  understands  how  each  step  follows  from  the  pre- 
ceding one. 

1.  The  segments  joining  the  mid-points  of  the  sides  of  an 
isosceles  triangle  form  an  isosceles  triangle. 

Analysis: 
I.  To  prove  AXYZ  isosceles,  prove  XZ=XY. 

IL    "       "      ZZ=Z7,  prove 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 

2.  How  would  Ex.  1  have  read  if  A  ABC  (Fig.  62) 
had  been  an  equilateral  triangle?    Give  proof. 

3.  If  the  segment  which  joins  the  vertex  of  a  triangle  with 
the  mid-point  of  the  base  is  perpendicular  to  the  base,  the  triangle 
is  isosceles. 

4.  In  Fig.  63,  CX  is  a  perpendicular  bisector  of  AB. 
C  moves  upward  along  the  ray  CX,  how  will  the 
sides  of  A  ACS  change?    What  are  the  upper  and 
lower  limits  to  these  sides?     No  proof  is  needed.  . 

5.  If  the  segment  which  bisects  the  vertex  angle 
of  a  triangle  is  perpendicular  to  the  base,  the  tri- 
angle is  isosceles. 

6.  In    Fig.    64,     CX    bisects     ZACB. 
If  the   segment   AB   moves  to   the   right  . 
and  to  the  left  along  the  ray  CX  but  is 
always  perpendicular  to  CX,  how  will  the  C-* 
sides    of    ACBA    change?    What    is    the 
lower  limit  to  the  length  of  these  sides?  .^j^ 
What   is   the   upper   limit?     No    proof    is              Fig.  64 
needed. 


CONGRUENT  TRIANGLES 


41 


7.  Given  the    equilateral  AABC   with   equal    segments  AX, 
BY,  and  CZ  measured  off  on  the  sides  as  shown  in 
Fig.  65,  prove  that  AXYZ  is  equilateral. 

8.  Given  the  isosceles  AABC  with  0  the  mid-point 
of  the  base  AB.  The  equal  sides  CA  and  CB  are 
extended  beyond  the  base  so  that  AE  =  BF.  Prove 
that  OE  =  OF. 


Fig.  65 


9.  Frames  of  various  shapes  can  be  made  by  fastening  together 
pieces  of  wood  or  iron  as  shown  in  Fig.  66.  Could  you  change 
the  shape  of  any  of  these  figures  by  pressing  upon  opposite  sides 
or  vertices?    Which  ones  could  be  so  changed? 


Fig.  66 

10.  How  are  roof  trusses,  elevated  train  structures,  bridges, 
and  the  like  made  rigid? 

11.  A  segment  drawn  from  the  center  of  a  circle  to  the  mid- 
point of  a  chord  is  perpendicular  to  the  chord.  Is  there  any 
special  position  of  the  chord  for  which  the  exercise  and  its  proof 
have  no  meaning? 

12.  Show  how  the  instrument  shown  in  Fig.  67 
may  be  used  as  a  leveling  device.  The  pieces  are 
so  fastened  together  that  AB  =  AC.  AD  is  a 
plumb  line.  How  must  the  framework  be  placed  if 
BC  is  level?  This  instrtmient  is  said  to  be  very 
ancient.* 


Fig.  67 


Suggestion.      Notice  that  a  level  line  is  one  that  is  perpendicular 
to  a  plumb  line. 

13.  Draw  any  triangle  ABC.     Find  X,  the  mid-point  of  AB. 
Join  CX  and  extend  so  that  XY  =  CX.    Prove  AC  =  BY. 
•See  D.  E.  Smith.  The  Teaching  of  Geometry,  p.  178. 


42 


PLANE  GEOMETRY 


Fig.  68 


14.  Fig.  68  shows  a   common   form   of   truss.     The  king-rod 
CO  joins    C   with   the  mid-point  of  AB  in   the 
isosceles  AABC.     The  braces  OE  and  OF  join  O        ^^ 
with   E  and   F,  the  mid-points  of  ^C  and   CB 
respectively.     Prove  that  OE  =  OF  and  that  CO 
is  perpendicular  to  AB. 

The  segment  joining  any  vertex  of  a  triangle  with  the 
mid-point  of  the  opposite  side  is  called  a  median  of 
the  triangle.  In  Ex.  14  the  king-rod  is  the  median  of 
the  isosceles  triangle  that  forms  the  truss. 


15.  In  Fig.  09,  AD  and  BC  are  ±  ^^  at  yl 
and  B  respectively.  If  AD  =  BC,  prove  that 
AC  =  BD. 


16.  Given  ABC,  an  isosceles  triangle  with  the 
equal  sides  extended  through  the  vertex  to  X  and 
Y.  AX  and  BY  are  ±  AB  at  points  A  and  B.  Prove 
that  AX  =  BY  (Fig.  70). 

Fig.  70 

17.  Given  the  isosceles  AABC  with  the  equal  sides  CA  and  CB 
extended  beyond  the  base  so  that  AX  =  BY,  prove  that  AY  =  BX. 

18.  In  Fig.  71,  ABC  is  an  isosceles  triangle 
with  AC  =  CB.  If  0  is  the  mid-point  of  AB, 
and  AC  and  BC  are  extended  so  that  CE  =  CD, 
prove  that  OE^OD. 

19.  If  any  point  0  in  the  bisector  of  the  vertex  angle  of  an 
isosceles  triangle  be  joined  to  the  extremities  of  the  base  AB, 
AAOB  is  isosceles. 

20.  The  segments  drawn  from  the  extremities  of  the  base  of 
an  isosceles  triangle  to  the  mid-points  of  the  opposite 

sides  are  equal. 

21.  In  Fig.  72,  ABC  is  an  isosceles  triangle.  CX 
and  CY  are  drawn  so  that  Z1=Z2.  Prove  that 
AXCY  is  isosceles. 


CONGRUENT  TRIANGLES  43 

22.  In  Fig.  73,  if  ABC  is  given  an  isosceles  triangle  and  CX 
and   CY   are   drawn   so   that    Z  1  =  Z  2,    prove    that  £ 

ACXY  is  isosceles. 


23.  The  three   medians  of  an   equilateral  triangle 

are  equal.  '^  '*      ^ 

^  Fig.  73 

24.  The  segments  that  bisect  the  base  angles  of  an  isosceles 
triangle  and  are  terminated  by  the  equal  sides  are  equal. 

25.  Draw  several  figures  for  Ex.  24.  Let  the  diff'erent  triangles 
have  the  same  base  but  legs  of  different  lengths.  Are  there  any 
limits  to  the  angle  that  the  bisector  of  one  of  the  base  angles 
makes  with  the  base?     No  proof  is  needed. 

26.  The  segments  that  bisect  the  angles  of  an  equilateral 
triangle  and  are  terminated  by  the  opposite  sides  are  equal. 

27.  If  from  the  ends  of  the  base  of  an  isosceles  triangle  seg- 
ments are  drawn  making  equal  angles  with  the  base  and  termi- 
nated by  the  opposite  sides,  these  segments  are  equal. 

REVIEW  DIAGRAMS 

53.  Review  the  proofs  of  Ths.  3  and  4  by  means  of  the 
following  scheme  and  diagram: 

Theorem  4 
If  three  sides  of  one  triangle  arc  equal  to  three  sides  of  another, 
the  triangles  are  congruent. 

1 1 \ 1 1 

Hyp.              Def.  iaoBcelea                          Th.3  Aa.go  Th.l 

triangle  | 

Fig.  74 

The  theorem  to  be  proved  is  written  down  in  full  with  a 
horizontal  line  below  it  (Fig.  74).  Below  the  various  ver- 
tical lines  shown  are  written  the  references  to  the  authorities 
on  which  the  proof  depends.  Some  of  these  references 
will  be  theorems.  These  theorems  (2  and  3  above)  should 
then  be  proved  and  the  references  to  the  authorities  used 
written  in  as  for  Th.  4.  This  process  should  be  con- 
tinued until  only  definitions  and  assumptions  remain. 


44  PLANE  GEOMETRY 

EXERCISES  THAT  MAY  INVOLVE  MORE  THAN  ONE  PAIR 
OF  CONGRUENT  TRIANGLES 

54.  Note.  When  exercises  are  selected  for  review  from  this  or  from 
the  next  section  it  is  often  well  to  require  the  proofs  for  all  theorems 
used  in  proving  the  exercises  assigned. 

L  Any  point  in  the  median  to  the  base  of  an  isosceles  triangle 
is  equally  distant  from  the  extremities  of  the  base. 

2.  Are    there    any    special  positions   of  the   arbitrary  point 
mentioned  in  Ex.  1   for  which  the  proof  given  does  not  hold? 

Q 

3.  In  Fig.  75,  CA  =  CB  and  CD  =  CE.  If  AD 
and  BE  intersect  at  O,  prove  that  CO  bisects  ZC. 

4.  In  Fig.  76,  CO  is  the  median  to  the  base  AB 
of  the  isosceles  AABC.  If  CX  =  CYy  prove  that  CO 
bisects  ZXOY. 

5.  Investigate    the    case,    Ex.  4,   in    which    X  ^ 
and  Y  are  on  BC  and  AC  extended.  ^^^-  ^^ 

6.  In  Fig.  77,  AC=-AD,   BC  =  BD,   AX=^AY. 
Prove  that  BX  =  BY. 

7.  Investigate  the  case,   Ex.    6,    in    which   X 
and  Y  are  on  AC  and  AD  extended.  Fig  77 

8.  In  Fig.   78,   CO    ±  line  h,    Rays  are  drawn  from  point 
C   so   that    Z 1  =  Z  2,  intersecting  line  h  at  points 
A  and  B.     A  and  B  are  joined  with  any  point  in  OC 
extended.     Prove  that  AABD  is  isosceles. 

9.  Investigate  the  case,  Ex.  8,  in  which  D  is  in 
CO  extended. 

10.  In  Fig.  79,  CA  =  CB  and  AD  =  BD.  If  X 
is  any  point  in  DC  extended,  prove  that  AX  =  BX. 

11.  Investigate  the  case,  Ex.  10,  in  which  X 
is  between  C  and  D.  Fig.  79 


CONGRUENT   TRIANGLES 


45 


Fig.  80 


12.  In  Fig.  80,  CD  is  any  segment.      Zl=  Z2  and  Z3=  Z4. 
Prove  that  CD  is  a  perpendicular  bisector  of  AB. 


13.  If,  in  Fig.  80,  A  ABC  is  constructed  isosceles 
and  Z>,  any  point  in  the  bisector  of  Z.C,  is  joined  to 
A  and  B,  DA  and  DB  make  equal  angles  with  AB. 

14.  Construct  the  figures  for  Exs.  12  and  13  with 
C  and  D  on  the  same  side  oi  AB. 

15.  In  Fig.  81,    ^ABC  is  isosceles.      CO  is  the 
median  to  the  base   AB.     If   Z 1  =  Z 2,  prove  that  1 
CX  =  CV  and  that  OX  =  OY. 


16.  In   Fig.    82,     AC  =  BD   and    AD  =  BC. 
Prove   that    AAOB   is   isosceles. 


Fig.  82 


17.  In  Fig.  83,  A  ABC  is  equilateral,  CE  = 
BD  =  AF.     Prove  that  AXZY  is  equilateral. 

18.  In  Fig.  83,  if  A  ABC  is  equilateral  and 
Z1=Z2=Z3,  prove  that  AXYZ  is  equi- 
lateral. Fig.  83 


19.  Prove  that  the  base  angles  of  an  isosceles 
triangle  are  equal  by  means  of  the  construction 
shown  in  Fig.  84.  A  ABC  is  the  isosceles  A. 
CX  =  CY. 

Analysis: 

To     prove     Z1=Z2,    prove   that     /.YAC=  /.CBX    and    that^ 
/.YAB^  AABX. 

Note.  This  method  of  proving  the  theorem  is  given  in  Euclid's 
Elements.  In  the  thirteenth  century  the  students  of  Oxford,  England, 
nicknamed  this  theorem  "elefuga" — that  is,  "the  flight  of  the 
wretched" — because  most  of  them  found  it  so  difficult  that  few  cared 
to  study  the   subject  further.     Two  or  three  hundred   years  later 


46 


PLANE  GEOMETRY 


the  students  called  it  "pons  asinorum,"  or  "the  bridge  of  asses." 
Very  little  is  known  concerning  Euclid  himself.  He  was  a  Greek  who 
lived  and  taught  in  Alexandria  about  300  B.C.  Some  of  the  earlier 
students  of  geometry  kept  the  results  of  their  studies  secret.  Euclid's 
Elements  was  a  great  advance  lipon  the  work  of  his  predecessors,  both 
in  arrangement  and  in  rigor.  It  is  said  that  Ptolemy  once  asked  him 
"if  there  was  in  geometry  any  shorter  way  than  that  of  the  Elements,'' 
and  he  answered  that  "there  was  no  royal  road  to  geometry."  Another 
story  told  of  him  is  that  some  one  who  had  begun  to  read  geometry 
with  EucHd  asked,  when  he  had  learned  the  first  theorem,  "But  what 
shall  I  get  by  learning  these  things?"  Euclid  called  his  slave  and  said, 
"Give  him  threepence  since  he  must  make  gain  out  of  what  he 
learns." 


20.  Show  that  the  distance  between  two  inaccessible 
as  A  and  B  on  opposite  sides  of  a  stream,  may  be  found  as 
(see  Fig.  85): 

Set  a  stake  at  C,  sighting  it  in  line  with  AB. 
BC  may  be  any  convenient  distance.  Take  D  any 
point  from  which  A,  B,  and  C  are  visible.  Sight 
E  in  line  with  D  and  C,  making  ED  =  DC.  Sight  F 
in  line  with  D  and  B,  making  FD  =  DB.  Sight 
G  so  that  it  will  be  in  line  with  F  and  E  and  also  ^^^ 
in  line  with  D  and  A.  What  line  should  be  measured 
the  distance  .45? 


points, 
follows 


0 
.  85 
to  find 


21.  Suppose  (Fig.  86)  that  P  represents  a  fence  post  on  one 
side  of  a  stream  and  line  /  a  fence  on  the  other  side  of  the  stream. 
A  fence  is  to  be  built  in  line  with  P  and 
perpendicular  to  fence  /.  Show  that  the 
following  method  may  be  used: 


Fig.  86 


Let  P  be  the  point  and  /  the  line. 
At  any  point  A  in  the  line  /  construct  a 
perpendicular  to  line  /  and  set  stakes 
making  AB  =  AC.  On  line  /  sight  D 
in  line  with  P  and  B,  and  E  in  line  with  C  and  P.  Then  sight 
F  in  line  with  D  and  C  and  at  the  same  time  in  line  with  E 
and  B.  PF  is  1.  I  and  S  is  the  point  where  the  ±  cuts  the 
line  /. 


CONGRUENT  TRIANGLES  47 

EXERCISES  INVOLVING   PROPERTIES    OF   AND    TESTS   FOR 

CONGRUENT  TRIANGLES  AND  CONSTRUCTION 

OF  TRIANGLES 

66.  1.  Corresponding  medians  of  congruent  triangles  are  equah 

Note.  Two  proofs  may  be  given:  one  by  means  of  congruent  tri- 
angles and  one  by  superposition. 

2.  Two  triangles  are  congruent  if  two  sides  and  the  median 
to  one  of  these  sides  are  ecjual  respectively  to  two  sides  and  the 
corresponding  median  of  the  other. 

3.  Construct  A  ABC  so  that  AB  =  7  cm.,  AC  =10  cm.,  and 
the  median  to  AC  =  4:  cm. 

Suggestion.  Draw  any  triangle  ABC  and  the  median  to  the  side  A  C. 
Write  the  numbers  7,  10,  and  4  on  the  proi)er  segments.  Look  at  the 
figure  until  it  is  evident  to  you  wliich  segments  must  be  constructed 
first  in  order  that  each  may  have  the  required  lenjiith. 

t4.  Construct  a  triangle,  given  two  sides  and  an  angle  opposite 
one  of  these  sides. 

Suggestion.  Let  a  and  h  be  the  given  sides  and  ZA  he  opposite 
side  a.  Show  how  Fig.  87  is  constructed.  DescTil)e  changes  in  the 
data  given  that  will  alter  the  results,  using 
the  following  outline: 

I.  Let  ZA  be  an  acute  angle. 
II.  Let  ZA  be  a.  right  angle. 
III.  Let  ZA  be  an  obtuse  angle. 

In  each  case  start  with  side  a  longer  than  ^^'" 

side  b,  then  suppose  side  a  to  decrease  gradually  and  note  results. 
Note.     This  problem  is  used  in  trigonometry  and  surveying. 

5.  Construct  an  isosceles  triangle  so  that  one  leg  shall  be 
9.4  cm.  and  the  median  to  that  leg  shall  make  with  that  leg  an 
angle  of  22K°. 

6.  Construct  A.-li5C  so  that  AB  =  H.6  cm.,  5C  =  6.5  cm.,  and 
the  median  to  AB  makes  with  AB  an  angle  of  45°. 

7.  The  bisectors  of  corresponding  angles  of  congruent  triangles 
are  equal. 

8.  Constmct  an  isosceles  triangle  so  that  one  base  angle  shall 
be  H  of  a  right  angle  and  one  leg  5  cm. 


CHAPTER  III 
Parallels,  Perpendiculars,  Angles,  Angle-Sums 

INTRODUCTORY 
PRELIMINARY  THEOREM:   TEST  FOR  UNEQUAL  ANGLES 

56.  If  one  side  of  a  triangle  is  extended,  an  angle  is  formed 
which  is  called  an  exterior  angle  of  the  tri- 
angle. Thus  in  Fig.  88,  AB  is  extended.  Z  1 
is  an  exterior  angle  of  AABC,  The  interior 
angle  numbered  2  is  adjacent  to  Z  1 ;  Z  3  and 
Z  4  are  the  non-adjacent  interior  angles. 

Exercise.     How  many  exterior  angles  has  a  triangle? 

57.  We  may  add  the  following  to  the  list  of  general 
assumptions : 

As.  28.  If  one  angle  or  segment  is  greater  than  a  second 
and  the  second  is  equal  to  or  greater  than  a  third,  then  the 
first  is  greater  than  the  third. 

As.  29.  The  whole  is  greater  than  any  of  its  parts. 

58.  Theorem  8.  An  exterior  angle  of  a  triangle  is  greater 
than  either  of  the  non-adjacent  interior  angles. 


Hypothesis:     In  AABC  the  side  AB  is  extended,  forming 
the  exterior  Z. 
Conclusion:     Zl  >  ZCor  /.BAC. 

48 


PARALLELS  AND  ANGLES 


49 


Fig.  90 


Analysis  and  construction: 

A.  I.  To  prove  Zl  >  ZC,  prove  part  of  Zl=  ZC 

II.    "        "     part  of  Z1=ZC,  bisect  CB  at  D,  join 
AD,   and   extend,  making  DE  =  AD.    Join  EB 
and  prove  Z 2  =  ZC. 
III.  To  prove  Z2=  ZC,  prove  ADBE  ^  AADC. 

B.  I.    "       "      Zl    >     ZBAC,   extend    CB   and   prove 

Z5  =    Zl  and  Z5  >   ZBAC.  c 

Let  the  pupil  give  the  proof. 

Ex.  1.  How  many  illustrations  can  you  find  in 
Fig.  90  of  an  exterior  angle  of  a  triangle?  Show 
how  Th.  8  applies  in  each  case. 

Ex.  2.  In  Fig.  90,  0  is  any  point  inside  AABC.    Prove  that 
ZAOB>ZC. 

TRANSVERSALS  AND  ANGLES 

59.  When  two  straight  Hnes  are  crossed  by  a  third  straight 
line,  various  angles  are  formed  which  have  special  names. 

Thus,  in  Fig.  91: 

Zc,   Zd,  Zw,  and  Zx  are  interior 

angles. 
Za,  Zb,  Zjy  and  Zz  are  exterior 

angles. 
Zc  and    Zx,   also    Zd  and    Zw, 

are  alternate  interior  angles. 
Za  and   Zz,  also   Zh  and   Zy,  are  alternate  exterior 

angles. 
Zaand  Zw,  Z6and  Zx,  Zc  and  Z;y,  also  Zd  and  Zz, 

are  corresponding  angles. 

Exercise.  In  Fig.  92,  name  8  pairs  of 
alternate  interior  angles,  16  pairs  of  corre- 
sponding angles,  and  8  pairs  of  alternate  ex- 
terior angles.  How  many  pairs  of  supple- 
mentary adjacent  angles  are  there?  Fig.  92 


Fig.  91 


50  PLANE   GEOMETRY 

PARALLELS 

60.  Lines  in  the  same  plane  that  do  not  intersect  however 
far  we  may  follow  them  are  called  parallel  lines.  Two 
arbitrary  straight  lines  in  the  same  plane  will  ordinarily 
intersect  and  determine  a  point.  Two  parallel  lines  do 
not  determine  a  point.  This  definition  is  the  fundamental 
test  for  parallels.  Five  other  tests  for  parallels  are  con- 
tained in  the  group  that  follows.  Th.  9  is  the  fundamental 
theorem  of  the  group. 

Exercise.  Find  in  the  room  in  which  you  are  sitting  illustra- 
tions of  parallel  lines  and  of  lines  which  do  not  intersect  and  yet 
are  not  parallel. 

FUNDAMENTAL  THEOREM:     TEST  FOR  PARALLELS 

61.  Exercise.  Construct  two  straight  Hnes  cut  by  a  third 
straight  line  so  that  the  alternate  interior  angles  are  equal. 

Theorem  9.  If  two  straight  lines  in  the  same  plane  are 
cut  by  a  third  straight  line  so  that  the  alternate  interior 
angles  are  equal,  the  two  straight  lines  are  parallel. 


Fig.  93 

Hypothesis:     Lines  a  and  b  are  cut  by  Hne  c  and  Z  1  =  Z  2. 

Conclusion:     Line  a  \\  line  6. 

Analysis: 
I.  To  prove  line  a  \\  line  b,  show  that  Hne  a  and  Hne  b 
cannot  meet  either  on  the  right  or  on  the  left. 

IL  To  prove  that  line  a  cannot  meet  line  b,  show  that  if 
line  a  met  line  b,  an  exterior  angle  of  a  triangle 
would  be  equal  to  an  opposite  interior  angle. 


PARALLELS  AND  ANGLES  51 

Proof: 

STATEMENTS  REASONS 

I.  a.  Line  a  might   meet  L  a.  Supposition, 

line  h  on  the  right. 

6.    Z  1  would  be  greater  6.  Why  ? 
than  Z2. 

c.  Zl  =  Z2.  c.  Given. 

d.  :.  line  a  does  not  d.  Supposition  a 
meet  line  b  on  the  leads  to  a  contra- 
right,  diction. 

IL  a.  Line   a  might   meet  II.  a.  Supposition, 

line  b  on  the  left. 
Let  the  pupil  complete  the  proof. 

62.  The  ordinary  direct  synthetic  proof  is  explained  in 
§46  and  has  been  used  for  theorems  and  exercises  in 
chapter  ii.  The  proof  used  for  Th.  9  is  an  indirect  proof. 
In  technical  terms  it  is  called  proof  by  reductio  ad  absurdum. 
This  is  a  Latin  phrase.     What  is  its  hteral  translation? 

For  such  proofs  we  must : 

1.  Determine  all  the  possible  cases  obtained  by  contra- 
dicting the  given  conclusion. 

2.  Then,  since  either  the  conclusion  or  one  of  the  con- 
tradictory statements  must  be  true,  we  must  eliminate  all 
but  one  of  these  by  proving  them  absurd. 

Proofs  of  this  character  are  very  common,  not  only  in 
mathematics,  but  in  all  argument.  Their  validity  depends 
among  other  things  upon  the  presentation  of  all  possibilities. 

The  proofs  for  many  of  the  theorems  and  exercises  that 
follow  are  clearer  if  expressed  in  algebraic  notation.  In 
some  cases  the  solution  of  an  equation  is  necessary.  In 
other  cases  the  use  of  algebraic  manipulations  and  iden- 
tities are  required  without  the  solution  of  an  equation. 
Be  definite  and   accurate. 


52  PLANE   GEOMETRY 

DEPENDENT  TESTS  FOR  PARALLELS 

63.  Theorem  10.  If  two  straight  lines  in  the  same  plane 
are  cut  by  a  third  straight  line  so  that  one  pair  of  correspond- 
ing angles  are  equal,  the  two  straight  lines  are  parallel. 


Fig.  94 

Hypothesis:     Lines  a  and  b  are  cut  by  line  n  so  that 
Z1  =  Z2. 
Conclusion:     Line  a  \\  line  b. 
Analysis: 
I.  To  prove  a  \\  b,  prove  Z  2  =  Z  3. 

IL    '*       ''      Z  2  =  Z  3,  compare  Z  2  and  Z  3  with  Z  1. 
Let  the  pupil  give  the  proof. 

Ex.    1.     Prove  Th.  10  by  proving  that   Z4=Z5.      Use  sup- 
plements of  equal  angles. 

Theorem  11.  If  two  straight  lines  in  the  same  plane  are 
cut  by  a  third  straight  line  so  that  the  interior  angles  on  the 
same  side  of  the  transversal  are  supplements,  the  two 
straight  lines  are  parallel. 

Hypothesis:  Lines  a  and  b  are  cut  by  line  n  so  that 
Z24-Z4  =  2rt.  Z. 

Conclusion:     Line  a  \\  line  b. 
Analysis  (see  Fig.  94) : 
L  To  prove  a  \\  b,  prove  Z  2  =  Z  3. 

II.  "  "  Z  2  =  Z  3,  show  that  Z  2  and  Z  3  are  each 
supplements  of  Z  4. 

Let  the  pupil  give  the  proof. 


PARALLELS  AND  ANGLES  53 

Theorem  12.  Two  straight  lines  in  the  same  plane  per- 
pendicular to  the  same  straight  line  are  paralleL 

Suggestion.     Prove  the  alternate  interior  angles  equal. 

Ex.  2.     Would  Ths.  9,   10,   11,  and  12  be  true  if  the  phrase 
in  the  same  plane  were  omitted? 

CONSTRUCTION  OF  PARALLELS 

64.  Problem  6.  To  draw  a  straight  line  through  a  given 
point  parallel  to  a  given  straight  line. 

Show  that  the  problem  may  be  solved  by  using  Th.  9, 
Th.  10,  or  Th.  12.  Make  the  construction  by  each  method 
and  prove  it.  How  many  lines  may  be  drawn  fulfilling  the 
requirements  ? 

Exercise.  Show  how  to  solve  Prob.  6  by  paper  folding,  or  with 
a  ruler  and  a  card,  or  two  draftsman's  triangles. 

FUNDAMENTAL  ASSUMPTION  REGARDING  PARALLELS 

65.  As.  30.  Only  one  line  can  be  drawn  through  a  given 
point  parallel  to  a  given  line. 

DEPENDENT  TEST  FOR  PARALLELS 

66.  Theorem  13.  Two  lines  parallel  to  a  third  line  are 
parallel  to  each  other. 

Suggestion.     Prove  by  the  indirect  method. 

EXERCISES   INVOLVING   TESTS   FOR   PARALLELS 

67.  To  prove  two  lines  parallel,  we  must  prove  one  of 
the  following: 

1.  The  alternate  interior  angles  are  equal; 

2.  The  corresponding  angles  aYe  equal; 

3.  The. interior  angles  on  the  same  side  of  the  trans- 

versal are  supplements; 

4.  They  are  perpendicular  to  the  same  line; 

5.  They  are  parallel  to  the  same  line. 


54  PLANE   GEOMETRY 

Ex.  1.  If  two  straight  lines  in  the  same  plane  are  cut  by  a 
third  straight  line  so  that  the  alternate  exterior  angles  are  equal, 
the  two  straight  lines  are  parallel. 

Ex.  2.  If,  in  Fig.  95,  lines  h  and  k  are  cut 
by  line  n so  that  Za4-Zft  =  2  vt.A,  prove  line  k  \\ 
line  h. 

Ex.  3.  If  any  two  segments  bisect  each  other, 
the  segments  joining  the  extremities  are  parallel. 

Ex.   4.     In  Fig.  96,  ABCD  is  a  four-sided 

figure  with  x  =  y    and  z  =  'w.     Prove  x  \\  y  and     

w  II  z.  ^         ^        ^ 

Fig.  96 

ANGLES  MADE  BY  PARALLELS  AND  TRANSVERSALS 
FUNDAMENTAL  THEOREM:  TEST  FOR  EQUAL  ANGLES 

68.  Theorem  14.    If  two  parallel  lines  are  cut  by  a  third 
straight  line,  the  alternate  interior  angles  are  equaL 


u 

aA 

h     A 

/ 

Fig. 

D          w 

95 

/ 

/ 

Fig.  97 

Hypothesis:     Line  h  ||  line  k  and   the  transversal  line  n 
cuts  lines  h  and  k. 

Conclusion:     Z 1  =  Z  2. 
Analysis  and  construction: 

I.  To  prove  Zl=  Z2,  construct  AB  so  that  /.BAC  = 

Z2  and  prove  Z.BAC=-  Zl. 
IL  To   prove    Z.BAC=  Z.\,  prove  BA    coincides   with 

line  h. 
III.  To  prove  line  BA  coincides  with  line  /t,  show  that 
BA  II  line  k.  and  line  h  ||  line  k. 


PARALLELS  AND  ANGLES 


55 


ProoJ: 


STATEMENTS 

REASONS 

^I. 

a.  Line  h  |1  line  k. 

L 

a.  Given 

b.  Line  AB  \\  line  k. 

h.  Why? 

c.  Line  AB  coincides  with 

c.  Only  one  line  can 

line  h. 

be  drawn  through 
a  given  point  par- 
allel to  a  given  line. 

IL 

Z  1  coincides  with  LB  AC. 

IL 

Why? 

in. 

Z1=Z2. 

III. 

Z 1  coincides  with 
an  angle  that  is 
equal  to  Z2. 

DEPENDENT  THEOREMS  CONCERNING  ANGLES  MADE  BY 
PARALLELS  AND  TRANSVERSALS 

69.  Theorem  15.    If  two  parallel  lines  are  cut  by  a  third 
straight  line,  the  corresponding  angles  are  equal. 


Fig.  98 

Hypothesis:     Line  k  ||  line  k,  and  line  n  cuts  lines  h  and  k. 
Conclusion:     Z  1  =  Z  2. 

Analysis:  To  prove  Z  1  =  Z2,  compare  Z 1  with  Z3  and 
Z2  with  Z3. 

Let  the  pupil  give  the  proof. 

Theorem  16.  If  two  parallel  lines  are  cut  by  a  third 
straight  line,  the  interior  angles  on  the  same  side  of  the 
transversal  are  supplements  of  each  other. 

Suggestion.     To  prove  that   Z2  is  the  supplement  of   Z4,  prove 
that  Z2  is  equal  to  an  angle  that  is  the  supplement  of  Z4. 


56  PLANE   GEOMETRY 

APPLICATION  OF  PARALLELS  TO  TEST  FOR  PERPENDICULARS 

70.  Theorem  17.    A  line  which  is  perpendicular  to  one 
of  two  parallels  is  perpendicular  to  the  other. 


k 

n 

h 

2 

Fig.  99 

Hypothesis:     Line  h  \\  line  k  and  line  n  J_  line  h. 

Conclusion:     Line  n  ±  line  k. 

Analysis: 

I.  To  prove  line  n  ±  line  k,  prove  Z 1  =  a  rt.  Z . 
II.    "        "     Zl  =  art.Z,  compare  Zland  Z2. 

The  proof  is  left  to  the  pupil. 

Ex.  1.  Prove  Th.  13  by  constructing  a  line  perpendicular  to 
one  of  the  given  lines  and  proving  it  perpendicular  to  the  others. 
Use  Ths.  17  and  12. 

Ex.  2.     Prove  Th.  13  by  drawing  a  transversal  and  using  Th.  9. 

71.  If  Ths.  9  and  14  are  compared,  it  will  be  seen  that 
they  are  related  to  each  other  in  a  peculiar  way.  In 
Th.  14  two  straight  lines  are  given  parallel  to  prove  that 
the  alternate  interior  angles  are  equal.  In  Th.  9  thfe  alter- 
nate interior  angles  are  given  equal  to  prove  that  the  two 
lines  are  parallel.  Th.  14  is  Th.  9  turned  around.  If  two 
theorems  are  so  related  that  the  hypothesis  and  conclusion 
of  one  are  respectively  the  conclusion  and  the  hypothesis  of 
the  other,  each  theorem  is  called  the  converse  of  the  other. 
If  a  theorem  is  proved  true,  its  converse  may  or  may  not 
be  true.  The  truth  of  the  converse  must  not  be  assumed; 
it  must  be  proved. 

Ex.  1.  State  and  investigate  the  truth  of  the  converse  of  the 
theorem:  If  two  triangles  are  congruent,  the  angles  of  one  are 
equal  respectively  to  the  angles  of  the  other. 

Ex.  2.     State  the  converse  of  Th.  15  and  of  Th.  16. 


PARALLELS  AND   ANGLES  57 

EXERCISES  INVOLVING  ANGLES  MADE  BY  PARALLELS 

J2.  L  If,  in  Fig.  100,  h  \\  k  and  Z  6  =  27°  30',  find  the  number  of 
degrees  in  each  angle  formed. 

2.  If  two  parallel   lines  are  cut  by  a  third  h         a/i 
straight  line,  the  alternate  exterior  angles  are 
equal. 

3.  If,   in  Fig.   100,  line   h    \\    line  k,   prove  ^^^  ^^^ 
that     /.a-\- /.x  =  2    rt.A.    What    other   angles 

in  the  figure  can  be  proved  supplementary  in  the  same  way? 

4.  If,  in  Fig.  101,  Z6  =  44°,  find  the  number  of  degrees  in  each 
angle  of  the  figure. 

5.  If,  in  Fig.  101,  Ax-  Zy  =  33°,  find 
the  number  of  degrees  in  each  angle  of  the 
figure. 

6.  If,   in  Fig.    101,   ZA  is   ^^  of  Z.y,  find  Fig.   101 
the  number  of  degrees  in  each  angle  of  the  figure. 

7.  In  Fig.  102,  /.BAC  is  any  angle.  0  is  any  point  on  the 
ray  AB.  If  the  point  X  starts  at  point  A 
and  moves  along  the  ray  AC  indefinitely, 
what  are  the  limiting  values  of  the  Z  XOA  and 
of  the  Z  BOX  ?  Investigate  the  case  in  which 
the  point  X  moves  from  A  along  AD. 

Fig.  102 

MISCELLANEOUS  EXERCISES 

73.  In  several  theorems  which  we  have  proved,  certain 
lines  were  constructed  and  used  which  were  not  given  in  the 
hypothesis.  Such  lines  are  called  construction  lines.  Their 
use  is  not  only  permissible  but  often  necessary.  Occasionally 
they  may  be  more  or  less  arbitrary.  When  such  lines  are 
located  defiftitely  care  should  be  taken  that  no  facts  are 
assumed  which  require  proof.  In  general,  two  points  or  one 
point  and  a  direction  locate  a  line.  For  methods  of  locating 
points,  lines,  rays,  or  segments  see  Ass.  1-8. 

1.  What  theorems  have  we  had  that  were  proved  by  the  aid 
of  construction  lines?    Tell  how  the  line  was  located  in  each  case. 


58  PLANE  GEOMETRY 

A  line,  ray,  or  segment  that  is  located  definitely  is  often 
called  a  fixed  line,  ray,  or  segment. 

'  The  use  of  construction  lines  in  the  following  exercises 
should  be  carefully  noted : 

2.  Lines  which  are  perpendicular  to  parallel  lines  are  parallel 
(Fig.  103). 

t3.  If  two  angles  have  their  sides  parallel  right 
side  to  right  side  and  left  side  to  left  side,  the     ~  I 

angles  are  equal  (Fig.  104).  Fig.  103 

Note.     The  right  side  of  an  angle  is  the  side  on  the  right  as  one 
stands  in  the  angle  and  faces  out. 

t4.  If    two    angles    have    their    sides      «//  / 

parallel  right  side  to  left  side   and  left      /  ^ — - —       /  / 

side  to  right  side,  the  angles  are  supple-         F'     '  ^'   / 

mentary.  Fig.  104 

5.  The  bisectors  of  a  pair  of  alternate  interior  angles  of 
parallel  lines  are  parallel. 

6.  If  through  the  vertices  of  an  isosceles  triangle  lines  are 
drawn  parallel  to  the  opposite  sides,  a  triangle  is  formed  which 
has  two  angles  equal. 

7.  How  would  Ex.  6  read  if  the  given  triangle  had  all  of 
its  angles  equal?     If  it  had  none  of  its  angles  equal? 

Give  proof. 

If,  in  Fig.  105,  line  h  \\  line  k  and  0  is  an  arbi-     ±. 


trary  point  between  the  parallels,  prove  Zb=  Za  -\-  Zc.     Fig.  105 

9.  If  a  four-sided  figure  has  both  pairs  of  opposite  sides 
parallel,  the  opposite  angles  are  equal. 

10.  If  a  four-sided  figure  has  both  pairs  of  opposite  sides 
parallel  and  one  angle  a  right  angle,  all  of  its  angles  are  right 
angles. 

11.  A  ray  parallel  to  the  base  of  an  isosceles  triangle  through 
the  vertex  bisects  the  exterior  angle  at  the  vertex. 

12.  What  would  be  true  in  Ex.  11  if  the  ray  parallel  to  the 
base  of  the  isosceles  triangle  cuts  the  sides  of  the  triangle  or  the 
sides  extended?     Give  proof. 


PARALLELS  AND  ANGLES  59 

13.  If  a  segment  between  two  parallel  lines  is  bisected,  any 
other  segment  between  the  parallels  and  through  the  point  of 
bisection  is  also  bisected  by  this  point. 

14.  In    Fig.     106,     AW\\BZ,    BX  =  AY,    and 
BZ  =  AW.    Prove    that    XZllYW. 


15.  In  Fig.    106,   if  AW  \\  BZ,  XZ\\  YW,   and 
BX  =  AY,  prove  that  BZ  =  AW.  Fig.  100 

ANGLES  IN  TRIANGLES 

FUNDAMENTAL  THEOREM 

74.  Theorem  18.     The  sum  of  the  interior  angles  of 
triangle  is  two  right  angles. 


Fig.  107 

Hypothesis:     A  ABC  is  any  triangle. 
Conclusion:     Zl+Z2+Z3  =  2rt.^. 
Analysis  and  construction: 
I.  To   prove    Zl  +  Z2  +  Z3  =  2   rt.Z,   compare    Z 1 

-h  Z  2  +  Z  3   with  angles  whose  sum  is  2  rt.  Z  . 
II.  Construct  XY  through  C  parallel  to  AB  and  com- 
pare Z  1,  2,  and  3  with  A  4,  5,  and  3  respectively. 

Let  the  pupil  give  the  proof. 

Note.  Th.  18  is  one  of  the  most  famous  theorems  of  geometry. 
It  is  supposed  that  the  ancient  Greeks  knew  that  it  was  true  for  equi- 
lateral and  for  isosceles  triangles  before  they  l<:ncw  that  it  was  true 
for  all  triangles.  The  proof  given  above  is  supposed  to  be  that  of 
Pythagoras  (about  500  B.C.)  and  may  be  one  of  the  earliest  proofs  for 
this  theorem. 

Ex.  1.     Can  you  verify  Th.  18  by  tearing  off  the  corners  made 
by  Al,  2,  and  3  in  Fig.  107  and  rearranging  them? 


60 


PLANE   GEOMETRY 


Ex.  2.    Are  there  any  other  ways  of  putting  in  construction 
lines  for  Th.   18  so  that   Z1  +  Z2  +  Z3 
may  be  compared  with  angles  whose  sum 
is  two  right   angles?     Give  proof.     (See 
Fig.  108.)  Fig.  108 

Ex.  3.     Find  the  number  of  degrees  in  the  third  angle  of  a 
triangle  if  the  other  two  angles  are: 

a.  40°,  56°  d,  59°  25',  58°  42' 

b.  29°,  58°  10'  e.  72°  16',  68°  42' 

c.  38°  40',  72°  18'  /.  58°  18',  79°  53' 


COROLLARIES:   VALUES  AND   COMPARISONS  OF  INTERIOR 
ANGLES  OF  TRIANGLES 

75.  Cor.  I.     Each  angle  of  an  equilateral  triangle  is  60°. 

Ex.   1.     Construct   angles   of   30°,-  15°,    75°,    7°  30',  67°  30', 
165°,  150°. 

Cor.  II.  If  two  angles  of  one  triangle  are  equal  respec- 
tively to  two  angles  of  a  second  triangle,  the  third  angles 
are  equal. 


Fig.  109 
Hypothesis:     A  ABC  and    AXYZ   have    ZA=  ZX  and 
ZB=ZY. 

Conclusion:     ZC=  ZZ. 
Analysis:     To  prove   ZC=  ZZ, 


Prove- 


ZA-\-ZB-}-ZC=ZX-\-ZY-}-ZZ, 


ZA+ZB==ZX-{-ZY. 

Ex.  2.  Segments  drawn  from  an  arbitrary  point  in  the  base 
of  an  isosceles  triangle  perpendicular  to  the  opposite  sides  make 
equal  angles  with  the  base. 


PARALLELS  AND  ANGLES  61 

Cor.  IIL    The  acute  angles  of  a  right  triangle  are  com- 
plements of  each  other. 

Ex.  3.  The  vertex  angle  of  an  isosceles  triangle  is  42°.  The 
perpendiculars  are  drawn  from  the  ends  of  the  base  to  the  opposite 
sides.  Find  the  number  of  degrees  in  the  angles  that  these  per- 
pendiculars make  with  the  base. 

THE   EXTERIOR   ANGLE   OF  A  TRIANGLE 

76.  Theorem  19.    The   exterior  angle   of  a  triangle   is 
equal  to  the  sum  of  the  two  non-adjacent  interior  angles. 


Fig.  110 

Hypothesis:     In  AABC  the  side  AB  is  extended,  forming 
the  exterior  Z  XBC. 

Conclusion:     Z  XBC=  Z  1  +  Z  2. 
Analysis  and  constrtcction: 
L  To  prove  ZXBC=  Z  1  +  Z2,  divide  ZXBC  into  two 
parts  and  compare  these  parts  with   Z  1  and   Z  2 
respectively. 
II.   .*.  draw  BY  through  B\\AC  and  compare  Z  4  with  Z  2 
and  Z5  with  Zl. 

Let  the  pupil  give  the  proof. 

Ex.  1.  An  exterior  angle  of  a  triangle  is  145°,  one  of  the 
opposite  interior  angles  is  62°.  Find  the  number  of  degrees  in 
each  of  the  angles  of  the  triangle. 

Ex.  2.  If  the  exterior  angle  at  the  vertex  of  an  isosceles 
triangle  is  128°,  find  the  number  of  degrees  in  each  of  the  angles 
of  the  triangle. 

Ex.  3.  Find  the  sum  of  the  exterior  angles  formed  when  the 
hypotenuse  of  a  right  triangle  is  extended  in  each  direction, 


62  PLANE   GEOMETRY 

EXERCISES  INVOLVING  THE  ANGLES  OF  A  TRIANGLE 

77.  1.  If  one  angle  of  a  triangle  is  double  the  second,  and  the 
third  is  double  the  first,  find  each  angle  of  the  triangle. 

2.  How  many  degrees  in  each  angle  of  an  isoscples  triangle 
if  the  vertex  angle  is  (1)  three  times  the  sum  of  the  base  angles? 
(2)  ^2  the  sum  of  the  base  angles?  (3)  \i  the  sum  of  the  base 
angles?  (4)  equal  to  the  sum  of  the  base  angles? 

3.  A  triangle  can  have  only  one  right  angle  or  one  obtuse 
angle.     May  a  triangle  have  a  right  angle  and  an  obtuse  angle? 

4.  If  the  vertex  angle  of  an  isosceles  triangle  is  42°,  find  the 
number  of  degrees  in  the  angles  at  the  intersection  of  the  bisectors 
of  the  base  angles.  Find  also  the  number  of  degrees  in  the  angles 
at  the  intersection  of  the  bisectors  of  the  exterior  angles  at  the 
base  of  the  triangle. 

5.  In  Fig.  Ill,  J\,ABC  is  isosceles.  AO  and 
OB  bisect  A  A  and  B  respectively.  If  the  legs 
CA  and  CB  are  made  to  increase  and  to  decrease 
in  length,  what  is  the  upper  limit  to  AAOB'i 
What  is  the  lower  limit?  ^i^-  ^ 

6.  If  the  vertex  angle  of  an  isosceles  triangle  is  62°,  find  the 
number  of  degrees  that  the  bisector  of  one  base  angle  makes  with 
the  opposite  side. 

7.  If  the  vertex  angle  of  an  isosceles  triangle  is  70°,  find  the 
number  of  degrees  in  the  angles  at  the  intersection  of  the  perpen- 
diculars drawn  to  the  equal  sides  from  the  extremities  of  the  base. 

8.  In  Fig.   112,  l\ABC  is  isosceles.     AX  and    BY  are  per- 
pendicular to   BC  and  AC  respectively.     If  the  legs  c 
AC  and  BC  are  made  to  increase  and  to  decrease  in 
length,  what  is  the  upper  limit  to   ZAOB  ?     What  is 
its  lower  limit? 

9.  If,  in  AABC,  ZA  is  4  times  ZC  and  ZB  is  3       ^ig.  112 
times  Z  C,   find  the  number  of  degrees  in  each  of  the  angles  of 
the  triangle. 

10.  Find  the  number  of  degrees  in  each  of  the  angles  of  an 
isosceles  triangle  if  each  base  angle  is  (1)  >!  the  vertex  angle; 
(2)  3  times  the  vertex  angle. 


PARALLELS  AND  ANGLES  63 

ANGLES   IN   POLYGONS 

78.  If  several  segments  are  joined  end  to  end  and  the 
free  end  of  the  last  is  joined  to  the  free  end  of  the  first,  the 
figure  formed  is  called  a  polygon.  The  segments  are  called 
the  sides  of  the  polygon;  the  common  end  points  of  the 
segments  are  called  the  vertices.  Segments  joining  any  two 
non-consecutive  vertices  are  called  diagonals.  The  sum  of 
the  sides  is  called  the  perimeter  of  the  polygon. 

A  polygon  is  said  to  be  convex  if  no  side  can  be  extended 
so  as  to  enter  the  polygon.  Otherwise  it  is  said  to  be 
concave  and  has  one  or  more  re-entrant  angles. 


Fig.  113 


In  Fig.  113  polygon  1  is  convex;  polygon  2  is  concave  with 
one  re-entrant  angle;  polygon  3  is  concave  with  two  re- 
entrant angles;  polygon  4  is  a  cross  polygon.  Hereafter 
unless  otherwise  stated  a  convex  polygon  is  intended. 

Polygons  are  named  according  to  the  number  of  sides: 

a  polygon  of  3  sides  is  called  a  triangle ; 

a  polygon  of  4  sides  is  called  a  quadrilateral; 

a  polygon  of  5  sides  is  called  a  pentagon ; 

a  polygon  of  6  sides  is  called  a  hexagon; 

a  polygon  of  7  sides  is  called  a  heptagon; 

a  polygon  of  8  sides  is  called  an  octagon ; 

a  polygon  of  10  sides  is  called  a  decagon; 

a  polygon  of  12  sides  is  called  a  duodecagon; 

a  polygon  of  15  sides  is  called  a  pentadecagon. 
Polygons  are  sometimes  called  by  their  English  instead  of 
by  their  Latin  or  Greek  names,  thus:  4-side,  7-side,  8-side, 
9-side,  etc. 


64  PLANE  GEOMETRY 

A  polygon  with  all  of  its  sides  and  all  of  its  angles  equal 
is  a  regular  polygon. 

THE  SUM  OF  THE  ANGLES  OF  ANY  POLYGON 

79.  Ex.  L     Find  the  sum  of  the  four  angles  of  a  quadrilateral. 
Ex.  2.    Find  the  sum  of  the  five  angles  of  a  pentagon  (Fig.  114). 
Analysis: 

To  find  the  value  of   ZA  + ZB+ ZC+ ^D-\- ZE,  i 

divide  the  polygon  into  triangles  whose  vertices  are  the         ,  //'  |\ 
vertices  of  the  polygon  and  find  the  sum  of  all  the  /    j    \ 

angles  of  all  the  triangles.      .*.  draw  the  diagonals  from  /      !    >^^ 

one  vertex  and  multiply  the  sum  of  the  angles  in  one         ^^^p 
triangle  by  the  number  of  triangles.  pj^   j  j^ 

Ex.  3.  Find  the  sum  of  the  angles  of  a  hexagon;  of  an  octagon; 
of  a  decagon. 

Theorem  20.  The  sum  of  the  iriterior  angles  of  a  poly- 
gon of  n  sides  is  2(n-2)  right  angles. 

Let  the  pupil  give  the  analysis. 

Proof: 

STATEMENTS 

1.  The    diagonals    divide    the    polygon   into ■  tri- 
angles. 

2.  The  sum  of   the  angles  of  each  triangle  is . 

3.  The  sum  of  the  angles  of  the triangles  is . 

Ex.  4.  By  substituting  the  proper  number  for  n  in  the  formula 
given  in  Th.  20  find  the  sum  of  the  angles  of  a  hexagon;  of  an 
octagon;  of  a  15-side;  of  a  16-side;  of  a  20-side;  of  a  24-side;  of  a 
32-side. 


Fig.  115 


Ex.  5.     Prove  Th.  20  by  means  of  the  constructions  shown  in 
Fig.  115. 


PARALLELS  AND  ANGLES  65 

Ex.  6.  How  many  degrees  in  each  angle  of  a  regular  octagon? 
of  a  regular  pentagon?  of  a  regular  decagon?  of  a  regular  12-side? 
of  a  regular  16-side?  of  a  regular  20-side?  of  a  regular  24-side?  of 
a  regular  n-side? 

Ex.  7.  Is  it  possible  to  have  a  regular  polygon  each  of  whose 
angles  is  108°?  150°?  144°?  128°?  160°?  If  such  polygons  are 
possible,  how  many  sides  would  there  be  in  each  case? 

Ex.  8.  Find  the  sum  of  the  interior  angles  of  a  peAtagon  that 
has  one  re-entrant  angle;  of  an  octagon  with  two  re-entrant  angles. 

Ex.  9.  How  many  regular  triangles  can  be  placed  adjacent  with 
their  vertices  at  the  same  point?  Will  the  space  about  the 
point  be  filled  exactly?    Why? 

Ex.  10.  Can  regular  hexagons  be  placed  with  their  vertices  at 
the  same  point  and  the  space  be  filled  exactly?  Why?  Can 
regular  quadrilaterals  be  so  placed?  regular  pentagons?  regular 
octagons?    Why? 

THE  SUM  OF  THE  EXTERIOR  ANGLES  OF  ANY  POLYGON 

80.  Ex.  1.     Find  the  sum  of  the  exterior  angles  of  a  pentagon. 
Analysis: 
To  find  the  sum  of  the  exterior  angles  of  ABCDE, 


subtract  the  sum  of  the  interior  angles  from  the  sum 

of  the  interior  and  exterior  angles  (Fig.  116).  ^^ -_ 

Let  the  pupil  give  the  proof.  Fig.  116 

Ex.  2.  Find  the  sum  of  the  exterior  angles  of  an  octagon;  of 
a  decagon;  of  a  12-side. 

Theorem  21.     The  sum  of  the  exterior  angles  of  a  poly- 
gon of  n  sides  is  four  right  angles. 

Let  the  pupil  give  the  analysis  and  the  proof. 

Ex.  3.  Is  it  possible  to  have  a  regular  polygon  each  of  whose 
exterior  angles  is  24°?  36°?  40°?  If  so,  how  many  sides  would 
these  polygons  have? 


66  PLANE  GEOMETRY 

MISCELLANEOUS  THEOREMS 
TEST  I  FOR  CONGRUENT  RIGHT  TRIANGLES 

81.  A  triangle  that  contains  a  right  angle  is  called  a  right 
triangle.  The  side  opposite  the  right  angle  of  a  right  triangle 
is  called  the  hypotenuse.  The  perpendicular  sides  are  called 
the  legs  of  the  right  triangle. 

Ex.  1.     Construct  a  right  triangle  with  the  hypotenuse  equal 
to  a  given  segment  and  one  acute  angle  equal  to  a  given  angle. 

Theorem  22.  Two  right  triangles  are  congruent  if  the 
hypotenuse  and  an  acute  angle  of  one  are  equal  to  the 
hypotenuse  and  an  acute  angle  of  the  other. 


Fig.  117 

Hypothesis:     In  AABC  and    AXYZ,   AC  =  XZ,    AA  = 
Z  X,  and  Z  B  and  Z  Y  are  rt.  A . 

Conclusion:     AABC  m  AXYZ. 

Analysis:     To  prove  AABC  ^  AXYZ,  prove  ZC=  ZZ. 

Let  the  pupil  give  the  proof.     Use  Cor.  II,  §75. 

Ex.   2.     Perpendiculars  dropped  from  the  mid-points  of  the 
equal  sides  of  an  isosceles  triangle  to  the  base  are  equal. 


Ex.  3.  Perpendiculars  from  the  mid-point  of  the 
base  of  an  isosceles  triangle  to  the  legs  are  equal. 

Ex.  4.  If,  in  Fig.  118,  CO  is  the  perpendicular 
bisector  of  AB,  what  are  the  limiting  values  of  the 
length  of  the  perpendicular  from  O  to  the  segment 
BX  as  X  moves  along  the  ray  OC? 


PARALLELS  AND  ANGLES 


67 


TEST  II  FOR  CONGRUENT  RIGHT  TRIANGLES 

82.  Ex.  1.     Construct  a  right  triangle  with  the  hypotenuse  and 
one  side  equal  respectively  to  given  segments. 

Theorem  23.  Two  right  triangles  are  congruent  if  the 
hypotenuse  and  a  side  of  one  are  equal  to  the  hypotenuse 
and  a  side  of  the  other. 


Fig.  119 

Hypothesis:  In  the  A  ABC  and  XYZ,  AC  =  XZ,  BC  = 
YZ,  and  Z  B  and  ZY  are  rt.  A  .    . 

Conclusion:     AABC  ^  AXYZ. 
Analysis  and  construction: 
I.  Toprove  AA^C  ^  AXYZ,  prove  ZA=  ZX, 
II.  To   prove    ZA  =  ZX,  place   AABC  so    that    BC 
falls  on  YZ,  B  on  Y,  C  on  Z,  and  A  opposite  X, 
and  prove  XYAZ  an  isosceles  triangle. 
III.  To  prove  XYAZ  a  triangle,  prove  XYA  a  straight 
line  (As.  19). 

Let  the  pupil  give  the  proof.     For  I  use  Th.  22. 

Cor.  If  a  perpendicular  is  erected  to  a  straight  line, 
equal  segments  drawn  from  the  same  point  in  the  per- 
pendicular cut  off  equal  distances  from  the  foot  of  the 
perpendicular. 

Ex.  2.  A  line  from  the  center  of  a  circle  perpendicular  to  a 
chord  of  the  circle  bisects  the  chord. 

Ex.  3.  Construct  an  isosceles  triangle,  given  one  leg  and  the 
perpendicular  from  the  vertex  to  the  base. 

Ex.  4.  Perpendiculars  drawn  from  an  arbitrary  point  in  the 
bisector  of  an  angle  to  the  sides  of  the  angle  are  equal. 


68  PLANE  GEOMETRY 

TESTS  FOR  ISOSCELES  TRIANGLES 

83.  Theorem  24.     If  two  angles  of.  a  triangle  are  equal, 
the  triangle  is  isosceles. 

The  analysis  and  the  proof  are  left  to  the  pupil. 

We  now  have  two  tests  for  isosceles  triangles. 
To  prove  a  triangle  isosceles,  prove  that 
L  Two  sides  are  equal,  or 
II.  Two  angles  are  equal. 

Of  these  two  tests  the  first  is  derived  from  the  definition 
of  an  isosceles  triangle  and  is  therefore  the  fundamental  one. 

Ex.  L     The  bisectors  of  the  base  angles  of  an  isosceles  triangle 
form  a  second  isosceles  triangle. 

Ex.  2.     The  bisectors  of  the  exterior  angles  at  the  base  of  an 
isosceles  triangle  form  a  second  isosceles  triangle. 

Ex.  3.     Construct  an  isosceles  triangle  whose  base  shall  be 
equal  to  a  given  segment  and  whose  vertex  angle  shall 
be  equal  to  a  given  angle. 


Ex.  4.     In  Fig.   120,  ABC  is  an  isosceles  triangle 
with  CA  =  CB.     If  CD  —  CE  and  AD  and  BE  intersect    a^ 
at  0,  prove  that  AAOB  is  isosceles.  Fig.  120 

Ex.  5.     In  Fig.  121,  AABC  is  isosceles.     AX  =  BY 
and  CZ  =  CW.     Extend   XZ  and    YW  to   meet   at  /  c^ 

0.     Prove  that  AXOY  is  isosceles.  zy^w 

AX  Y     B 

Ex.  6.     If,  in  Fig.   121,    AXOY   is   constructed       Fig.  121 
isosceles,  XA  =  YB  and  XZ=YW,  prove  that  AABC  is  isosceles. 

PROPERTY  OF  ISOSCELES  TRIANGLES 

84.  Theorem  25.  A  segment  from  the  vertex  of  an  isos- 
celes triangle  perpendicular  to  the  base  bisects  the  base 
and  the  vertex  angle. 

The  analysis  and  the  proof  are  left  to  the  pupil. 

For  other  properties  of  isosceles  triangles,  see  Th.  6  and  Th.  7. 


PARALLELS  AND  ANGLES 


69 


Fig.  122 


SUPPLEMENTARY   EXERCISES 

EXERCISES   INVOLVING   ANGLES   OF   POLYGONS 

85.  1.  In  Fig.  122,  ABCDE  is  a  convex  polygon  of  five  sides,  all 
of  whose  angles  are  obtuse.  The  sides  of  the 
polygon  are  extended  until  they  intersect,  form- 
ing the  star  polygon  shown.  Find  the  sum  of 
the  angles  in  the  points  of  the  star.  What 
would  be  the  sum  of  the  angles  in  the  points 
of  the  star  if  the  polygon  had  six  sides?  If  it 
had  eight  sides? 

2.  Suppose  the  convex  polygons  used  in  Ex.  1  were  regular, 
how  many  degrees  would  there  be  in  the  angle  at  any  point  of 
the  star  polygons  formed? 

3.  Fig.  123  shows  a  regular  triangle  with  each 
side  divided  into  three  equal  parts.  If  the  points 
are  joined  as  indicated,  prove  that  DEFGHK  is  a 
regular  hexagon. 

4.  Fig.  124  shows  a  kite  formed  of  two  regular 
triangles  with  the  same  base  AC.  If  the  sides 
are  bisected  and  the  points  joined  as  shown,  prove 
that  AEFCGH  is  a  regular  hexagon. 


5.  Fig.  125  shows  a  regular  hexagon  ABCDEF, 
If  alternate  sides  are  extended  in  both  directions  as 
indicated,  prove  that  a  regular  triangle  XYZ  is 
formed. 


Note.     Tiled   and    mosaic   floors   are   often    made   of   equilateral 
triangles   so   colored    that   various   patterns   are   formed.     Fig.    126 


XJLXJL 


^ 


Fig.  126 
represents  three   of  these  designs  containing  the  figures  used  in  the 
preceding  exercises.     The  use  of  equilateral  triangles,  squares,   and 
hexagons  for  tiles  probably  dates  back  to  the  ancient  Egyptians. 


70  PLANE  GEOMETRY 

MISCELLANEOUS   EXERCISES 

86.  Note.     Be  prepared  to  prove  the  theorems  on  which  each  of 
the  following  exercises  depends. 

A  perpendicular  from  any  vertex  of  a  triangle  to  the 
opposite  side  is  called  an  altitude  of  the  triangle. 

1,  Make  review  diagrams  for  Ths.  11,  15,  19,  and  25. 

2.  If  ABC  is  an  isosceles  triangle  and  if  the  perpendiculars 
A  Y  and  BX  drawn  from  the  extremities  of  the  base  AB\.o  the  equal 
sides  BC  and  ^C  intersect  at  0,  t\AOB  is  isosceles. 


3.  In  Fig.  127,  ^5  is  any  segment,  Zl=  Z2, 
and  AD  =  BC.     From  D  and  C  perpendiculars  are 
drawn  to  AB.     What    segments  and  angles   are  Fig.  127 
equal?    Why? 

4.  Investigate  the  case,  Ex.  3,  in  which  Zl  and  Z2  are 
obtuse  angles.     Give  proof. 

5.  If  the  ray  which  is  drawn  through  the  vertex  of  a  triangle 
parallel  to  the  base  bisects  the  exterior  angle  at  the  vertex,  the 
triangle  is  isosceles. 

6.  A  segment  drawn  from  an  arbitrary  point  in  the  bisector 
of  an  angle  to  one  side  of  the  angle  and  parallel  to  the  other  side 
forms  with  the  bisector  and  the  side  to  which  it  is  drawn  an 
isosceles  triangle. 

7.  In  Fig.  128,  BD\\  XY.  AB  bisects  ZX^IC  and  AD  bisects 
Z  CA  F.     Prove  that  BC  =  CD.  x_ 

8.  Construct   an   isosceles   triangle   with   the 
vertex  angle   Vs  of  a  right   angle   and  the  alti-  b  c" 
tude  4.7  cm.                                                                      ^^^-  ^28 

9.  In  Fig.  129,  ABC  is  an  isosceles  triangle 
with  D  any  point  in  ^C  extended.  From  D  a  per- 
pendicular is  drawn  to  AB  cutting  CB  at  E.  Prove 
that  CDE  is  an  isosceles  triangle. 

10.  Investigate  the  case,  Ex.  9,  in  which  the 
point  F  is  on  J  5  extended. 


PARALLELS  AND  ANGLES 


71 


U.  In  Fig.  130,  ABC  is  an  isosceles  right  triangle 
with  ZC  a  right  angle.  CD  is  perpendicular  from 
Cto  AB.    Construct  XY=YC  and  YZ=  YC. 


Note.  The  patterns  used  in  applied  design  are  usually  made  by 
repeating  at  regular  intervals  some  very  simple  figure  called  the  unit. 
Figs.  131  and  132  show  two  parquet  floor  patterns.  Show  how  Fig. 
130  in  whole  or  in  part  is  used  in  each  of  these  designs.  The  possi- 
bilities of  Fig.  130  as  a  design  unit  may  be  discovered  by  making  four 
or  eight  copies  of  the  figure,  coloring  the  spaces  to  suit  one's  fancy, 
cutting  the  figures  out,  and  fitting  them  together  in  various  ways. 


^^ 

m 

m 

Fig.  131 


Fig.  132 


ABC  is  an 


12.  Construct  the  fan  truss  shown  in  Fig.  133. 
isosceles     triangle.      AD  =  DC  =  CE  =  EB. 
AF  =  FD  =  DG  =  GC.     BK  =  KE  =  EH  =  HC.  <^^ 
Compare  LEDC  and   LDAC. 

A  l>  ^- 

13.  Construct  an  isosceles  triangle,  given  p^^,  jgg 

the  perimeter  equal  to  a  given  segment  and     From  a  roof  truss  dcsiKn 
one  base  angle  equal  to  a  given  angle. 

Suggestion.     Study  Fig.  133.      CD  +  CE+DE  is  to  be  equal  to  the 
given  segment  AB,  and  Z.EDC  is  to  be  equal  to  the  given  angle. 

14.  Construct   a  triangle,  given  the   perimeter  and   the   two 
base  angles. 

15.  Show  that  a  sailor  may  calculate  his  dis- 
tance from  a  lighthouse  as  follows:  In  Fig.  134 
C  represents  the  lighthouse  and  BA  the  direction  in 
which  the  boat  is  moving.  With  the  proper  instru- 
ments he  may  take  the  angle  CBA.  Then,  if  he 
notices  the  moment  when  the  angle  ( Z  CAD)  which  the  lighthouse 
makes  with  his  course  is  twice  Z  B,  he  can  find  the  length  of  CA, 
for  he  can  obtain  the  length  oi  AB  from  his  log.  (From  the  Final 
Report  of  the  National  Committee  of  Fifteen.) 


Fig.  134 


72  PLANE   GEOMETRY 

16.  In  surveying,  it  is  often  necessary  to  run  a  line,  as  CD 
(Fig.  135),  so  that  it  will  be  in  the  same 
straight  line  with  LA  but  on  the  othfer  side 
of  some  obstacle  to  vision,  such  as  a  house  ^^^ 

or  wood.     Show  that  the  following  method  Fig.  135 

will  give  the  desired  result:  Lay  off  the  Z  a  so  as  to  clear  the 
obstacle.  Take  AB,  a.  convenient  distance.  Lay  off  Zb  =  2Za. 
Make  BC=BA,     Lay  off  Zc=  Za. 

^    17.  Perpendiculars  from  the  ends  of  the  base  of  an  isosceles 
triangle  to  the  opposite  sides  are  equal. 

18.  In  any  AABC  perpendiculars  from  points  A  and  B  to 
the  median  to  the  side  AB  are  equal.  For  what  special  case  of 
the  arbitrary  AABC  is  the  proof  of  this  exercise  meaningless? 

19.  If  two  parallel  lines  are  cut  by  a  third  straight  line,  the 
bisectors  of  the  interior  angles  on  the  same  side  of  the  transversal 
are  perpendicular  to  each  other. 

20.  If  two  parallel  lines  are  cut  by  a  third  straight  line,  the 
bisectors  of  the  four  interior  angles  form  a  quadrilateral  with 
four  right  angles. 

21.  If  a  perpendicular  be  drawn  from  the  vertex  of  the  right 
angle  of  a  right  triangle  to  the  hypotenuse,  the  two  triangles  formed 
are  mutually  equiangular. 

22.  Construct  an  isosceles  triangle  with  altitude  4.3  cm.  and 
the  base  angles  each  H  of  a  right  angle. 

23.  Corresponding  altitudes  of  congruent  triangles  are  equal 
(Fig.  136). 

24.  Two  triangles  are  congruent  if  two  ^  g, 
sides  and  the  altitude  to  the  third  side 
of  one  are  equal  respectively  to  two  sides 
and  the  altitude  to  the  third  side  of  the 
other  (Fig.  136). 


B  D 

Fig.  136 


25.  Prove  Ex.  24  when   /.B  and   Z  £  (Fig.  136)  are  obtuse. 

26.  Construct   AABC  so  that  AC  =  10  cm.,  BC  =  7  cm.,  and 
the  altitude  on  AB  =  6  cm. 


PARALLELS  AND  ANGLES  73 

27.  Two  triangles  are  congruent  if  two  sides  and  the  altitude 
on  one  of  these  sides  in  one  triangle  are  equal  respectively  to  two 
sides  and  the  corresponding  altitude  of  the  other. 

28.  Construct  A  ABC  so  that  AB  =  7  cm.,  AC^b  cm.,  and 
the  perpendicular  from  C  to  ^45  =  3  cm. 

29.  The  line  bisecting  the  exterior  angle  at  the  vertex  of  an 
isosceles  triangle  is  parallel  to  the  base. 

30.  The  angle  made  by  the  bisectors  of  the  base  angles  of 
an  isosceles  triangle  is  equal  to  the  exterior  angle  at  the  base  of 
the  triangle. 

3L  If  ABD  is  an  isosceles  triangle  with  DA  and  DB  equal 
sides,  and  if  i4D  is  extended  through  D  to  point  C  until  DC  =  DA, 
prove  that  BC  is  perpendicular  to  AB. 

32.  If  an  exterior  angle  of  a  triangle  is  bisected  and  also  one 
of  the  interior  non-adjacent  angles,  the  angle,  made  by  the  two 
bisectors  is  ^  the  other  interior  non-adjacent  angle. 

33.  Construct  through  a  given  point  a  ray  that  shall  make  a 
given  angle  with  a  given  line.  Is  there  more  than  one  solution 
for  this  problem? 

34.  If  two  medians  of  a  triangle  are  extended  beyond  their 
bases  and  segments  are  taken  on  the  extended  lines  equal  to  the 
corresponding  medians,  the  points  thus  found  and  the  other  vertex 
of  the  triangle  are  on  a  straight  line  that  is  parallel  to  the  opposite 
side  of  the  triangle. 

35.  In  Fig.  137,  ABC  is  an  isosceles  triangle  with 
ZA  twice  ZC.  Find  the  number  of  degrees  in  each 
angle  shown  in  the  figure  ii  AX  bisects   ZA. 


Note.      Fig.    137    cannot   be    constructed    at    present 
without  a  protractor. 


Fig.  137 


36.  In  Fig.  138,  ABC  is  an  isosceles  triangle 
with  ZB  =  54°.  AX±CB  from  A .  Find  the  number 
of  degrees  in  Z  1  and  Z  2. 

37.  Construct  AABC  with  ZA=30°,  ZB  =  45°, 
and  the  perpendicular  from  C  to  AB  6.4  cm.  Pig.  138 


74 


PLANE   GEOMETRY 


t38.  In  each  of  the  figures  shown  in  Fig.  139,  lines  a  and  a' 
are  perpendicular  to  each  other,  also  lines  b  and  b'.  Prove  that 
Zab=  Za'b'. 

Suggestion.     Apply  Cor.  Ill,  §75. 


Fig.  139 


139.  Investigate  the  truth  of  the  statement  that  if  two  angles 
have  the  sides  of  one  perpendicular  respectively  to  the  sides  of 
the  other  the  angles  are  equal. 

40.  In  Fig.  140,  A  BCD  is  a  four-sided 
figure  with  its  sides  equal  and  its  angles  right 
angles.  AE  =  BF  =  BG  =  CH  =  etc.  EX  and  HY 
are  parallel  to  ^C;  XF  and  MW  are  parallel  to 
DB.  NW,  KZ,  GY,  and  LZ  are  similarly  drawn. 
Prove  that  ANOE,  EOF,  and  EXF  are  isosceles. 
How  many  isosceles  triangles  does  the  figure 
contain?  What  triangles  in  the  figure  are  congruent?  If  H  and  E 
are  joined,  prove  EH  ||  AC. 

41.  Find  the  relation  between  the  base  angles  of  two  isosceles 
triangles  if  the  vertex  angles  are  supplementary. 

42.  If  a  segment  meets  the  sides  of  an  isosceles  triangle  at 
equal  distances  from  the  vertex,  it  is  parallel  to  the  base. 

43.  If  the  sides  of  two  angles  are  parallel  right  side  to  right 
side  and  left  side  to  left  side,  the  bisectors  of  the  angles  are 
parallel.  / 

44.  Find  the  number  of  degrees  in  the  angles  at  the  intersec- 
tion of  the  bisectors  of  the  acute  angles  of  a  right  triangle. 

45.  Prove  that  a  convex  polygon  cannot  have  more  than  three 
obtuse  exterior  angles  or  more  than  three  acute  interior  angles. 


CHAPTER  IV 

Quadrilaterals 

SYMMETRY 

87.  We  have  seen  in  chapter  ii  that  under  certain  cir- 
cumstances two  figures  can  be  placed  one  upon  the  other  so 
as  to  coincide  exactly.  Moreover,  if  we  wish  to  prove  two 
segments  or  two  angles  equal,  we  often  look  for  two  triangles 
that  contain  these  segments  or  angles  and  try  to  prove  these 
triangles  congruent.  In  many  cases  we  can  make  one  of 
these  triangles  coincide  with  the  other  by  folding  the  figure 
along  some  line  in  the  figure,  or  by  rotating  part  of  the  figure 
about  some  one  point. 


No.  1 


Fig.  141  shows  four  figures  of  this  kind  which  we  have  had. 
If  Nos.  1  and  2  are  folded  along  the  line  AB,  the  two  parts 
will  coincide.  If  Nos.  3  and  4  are  rotated  about  point  O 
through  180°,  each  figure  will  coincide  with  its  original 
impression.  Figures  or  parts  of  figures  that  can  be  made  to 
coincide  in  either  of  these  ways  are  said  to  be  symmetric. 
Points,  lines,  segments,  or  angles  that  coincide  under  these 
circumstances  are  said  to  be  symmetric  to  each  other. 

Exercise.  Can  you  find  any  other  figures  in  chapter  ii  or 
chapter  iii  which  are  symmetric?  Can  you  find  any  figures  in 
these  chapters  which  are  not  symmetric? 

75 


76  PLANE  GEOMETRY 

DEFINITIONS   OF  AXIAL   SYMMETRY 

88.  A  figure  is  said  to  be  symmetric  with  respect  to  a 
line  as  an  axis  if  one  part  coincides  with  the 
remainder  when  it  is  folded  on  that  line  as  an 
axis  (Fig.  142). 

Two  figures  are  said  to  be  symmetric  with 
respect  to  a  line   as   an   axis  if  one  figure 
coincides  with  the  other  when  the  plane  in 
which  it  lies  is  folded  on  that  line  as  an  axis      > 
(Fig.  143). 

Such  a  figure  or  such  figures  are  said  to 
have  axial  symmetry.  Fig.  143 

THEOREMS  AND   EXERCISES   INVOLVING   AXIAL 
SYMMETRY 

89.  Theorem  26.  The  bisector  of  the  vertex  angle  of  an 
isosceles  triangle  is  an  axis  of  symmetry  of  the  triangle. 

Analysis:  ^ 

To  prove  that  CD  is  an  axis  of  symmetry  of 
A  ABC  (Fig.  144),  prove  that  AACD  will  coin-    ^     ^     ^ 
cide  with  ABCD  if  AABC  is  folded  on  CD  as      Fig.  144 
an  axis. 

Ex.  1.  The  end  points  of  a  segment  are  symmetric  with 
•respect  to  the  perpendicular  bisector  of  that  segment  as  an  axis. 

Ex.  2.     What  axes  of  symmetry  have  two  parallel  lines? 

Ex.  3.  Show  how  to  place  two  congruent  triangles  so  that 
they  are  symmetric  with  respect  to  a  side  of  one  as  an  axis. 

90.-  Theorem  27.  Two  polygons  are  symmetric  with 
respect  to  an  axis  if  the  vertices  of  one  are  symmetric  to 
the  corresponding  vertices  of  the  other. 

Ex.  1.  How  would  Th.  27  read  if  the  two  polygons  were 
parts  of  the  same  polygon? 

Ex.  2.  Show  how  to  construct  a  pentagon  symmetric  to  a 
given  pentagon  with  a  given  line  as  axis, 


QUADRILATERALS  77 

DEFINITIONS  OF  CENTRAL  SYMMETRY 

91.  A  figure  is  said  to  be  symmetric  with  respect  to  a 
point  as  a  center  if  one  part  of  the  figure  coin- 
cides with  the  remainder  when  it  is  rotated 


through  an  angle  of  180°  about  the  point  as  a 

fcenter  (Fig.  145).  Fig.  145 

Two  figures  are  said  to  be  symmetric  with  *?=^^^ 

respect  to  a  point  as  a  center  if  one  figure  ^^^  •/ 

coincides  with  the  other  when  it  is  rotated  /;\ 

through  an  angle  of  180°  about  the  point  as  ^\|^::i^ 

a  center  (Fig.  146).  Fig,  ^q 

Such  a  figure  or  such  figures  are  said  to  have  central 
symmetry.  These  definitions  give  us  the  following  test  for 
central  symmetry:  A  figure  is  symmetric  with  respect  to 
a  point  as  a  center  if  for  every  point  in  it  there  is  a  cor- 
responding point  so  situated  that  the  two  points  are 
sjrmmetric  with  respect  to  the  center.  Can  you  state  a 
similar  test  for  axial  symmetry? 

THEOREMS   AND   EXERCISES   INVOLVING   CENTRAL 
SYMMETRY 

92.  Ex.  1.     The  center  of  symmetry  of  two  points  is  the  mid- 
point of  the  segment  joining  the  two  points. 

Ex.  2.     Two  vertical   angles  are  symmetric  with   respect  to 
their  vertex  as  a  center. 

Ex.  3.     Find  a  center  of  symmetry  of  two  parallel  lines.     How 
many  such  centers  are  possible? 

Theorem  28.  Two  polygons  are  symmetric  with  respect 
to  a  center  if  the  vertices  of  one  are  symmetric  to  the  cor- 
responding vertices  of  the  other. 

Ex.  4.     How  would  Th.  28  read  if  the  two  polygons  were 
halves  of  the  same  polygon? 

Ex.  5.     Which  letters  of  the  alphabet  have  central  symmetry? 
Which  ones  are  symmetric  with  respect  to  an  axis? 


78 


PLANE  GEOMETRY 


RELATION  BETWEEN  AXIAL  AND  CENTRAL  SYMMETRY 

93.  Theorem  29.  Any  figure  that  has  two  axes  of  sym- 
metry at  right  angles  to  each  other  has  the  intersection  of 
the  axes  as  a  center  of  symmetry. 


v> 
Fig.  147 

Hypothesis:     P  is  a  point  on  any  figiire  which  is  symmetric 
with  respect  to  xx'  and  yy'  as  axes.     Axis  xx'  _L  axis  yy' . 

Conclusion:     0,  the  intersection  of  xx'  and  yy' ,   is   the 
center  of  symmetry  of  the  figure. 
Analysis: 
I.  To  prove  the  figure  s)mimetric  with  respect  to  point 
O,  prove  that  for  every  point  in  the  figure  there 
exists  a  point  symmetric  to  it  with  respect  to  0  as 
a  center. 
II.    .*.  let  P  be  any  point  in  the  figure  and  P'  be  sym- 
metric   to    P    with   respect    to   yy'    and    P"   be 
symmetric  to  P'  with  respect  to  xx'  and  prove  O 
the  mid-point  of  PP". 

III.  To  prove  O  the  mid-point  of  PP",  join  PO,  P'O,  P"0, 

and  prove  PO  =  P'V  and  POP"  a  straight  line. 

IV.  To  prove  PO  =  P"0,  prove  them  both  equal  to  P'O. 

V.  To  prove  that  POP"  is  a  straight  line,  prove  that 
Zi-1-Z2+Z3+Z4  =  2rt.  A. 

VI.  To  prove  that  Z  1+ Z2+ Z3+ Z4  =  2  rt.  A,  prove 
Z1=Z2,    Z3=Z4,  and   Z2-}-Z3  =  l  rt.    Z. 

Let  the  pupil  give  the  proof  in  full. 


QUADRILATERALS 


79 


Ex.  L  How  many  axes  of  symmetry  has  an  equilateral  tri- 
angle?    Has  it  a  center  of  symmetry? 

Ex.  2.  Prove  that  a  quadrilateral  with  its  four  sides  equal 
has  a  center  of  symmetry. 

Note.  Symmetric  figures  are  much  used  in  ornament.  Designers 
make  constant  use  of  the  idea  of  symmetry.  Illustrations  may  bo 
found  in  ornamental  windows,  wall  paper,  etc.  Symmetry  also  occurs 
in  nature,  as  in  snow  crystals,  flowers,  etc.  Fig.  148  shows  three  sym- 
metric encaustic  tile  designs.  Let  the  pupil  find  other  illustrations. 
How  might  kaleidoscopes  and  mirrors  be  used  by  designers? 


K^SZrasa 


^V/?f^1 


Fig.  148 


PARALLELOGRAMS 

DEFINITIONS 

94.  Ex.  1.  Construct  ZABC  =  (jO°.  Make  ^5  =  5.3  cm.  and 
BC  =  G.7  cm.  From  A  construct  AD  parallel  to  ^C.  From  C 
construct  CD  parallel  to  AB. 

Ex.  2.  Construct  a  quadrilateral  with  its  opposite  sides 
parallel.  Make  one  angle  45°  and  the  sides  that  include  the 
angle  3.8  cm.  and  5.9  cm.  respectively. 

A  quadrilateral  with  each  side  parallel  to  its  opposite  is 
called  a  parallelogram. 

In  Fig.  149,  X  and  z,  also  w  and 
y,  are  called  opposite  sides ;  w  and  x 
are  called  consecutive  sides;  A  A 
and  C  are  called  opposite  angles; 
A  A  and  D  are  called  consecutive  Fig.  149 

angles;  AC  and  BD  are  called  diagonals.  While  any  side 
may  be  considered  as  the  base,  x  and  z  are  usually  called 
the  bases. 


80  PLANE   GEOMETRY 

PROPERTIES    OF   PARALLELOGRAMS 

95.  The  fundamental  characteristic  of  parallelograms  is 
stated  in  the  definition,  namely:  The  opposite  sides  are 
parallel.  The  next  three  theorems  depend  directly  upon 
this  fact. 

Theorem  30.  Each  diagonal  of  a  parallelogram  divides 
it  into  two  congruent  triangles. 

Theorem  31.  The  opposite  sides  of  a  parallelogram  are 
equal. 

Theorem  32.  The  opposite  angles  of  a  parallelogram  are 
equal. 

Ex.  1.     Two  consecutive  angles  of  a  parallelogram  are  supple- 
mentary. 

Ex.  2.  The  sum  of  the  angles  of  a  parallelogram  is  four  right 
angles. 

96.  The  diagonals  of  a  parallelogram  differ  from  the 
diagonals  of  other  four-sided  figures  in  important  respects. 
Note  the  exercise  on  p.  81. 

We  shall  assume  that  the  diagonals  of  convex  quadri- 
laterals intersect. 

Theorem  33.  The  diagonals  of  a  parallelogram  bisect 
each  other. 


Analysts: 

I.  To  prove  that  the  diagonals  bisect  each  other,  prove 
that  AO  =  OC  and  that  BO  =  OD. 

IL  To  prove  I  ^^^^^,  prove  ADOC  ^  AAOB . 


QUADRILATERALS  81 

Exercise.     Fig.  151  shows  a  parallelogram,  a  convex    quadri- 
lateral,   and    a   concave    quadrilateral.     In  V    ~~7  /      1     i -i 

which  cases  do  the  diagonals  intersect?     In    \\    ^^     1  / / 

which  case  do  they  bisect  each  other?  v       ^^~^ 

Fig.  151 

97.  Theorem  34.  The  intersection  of  the  diagonals  of  a 
parallelogram  is  the  center  of  symmetry  of  the  parallelogram. 

CONGRUENCE   OF   PARALLELOGRAMS 

98.  The  OR  I'M  35.  Two  parallelograms  are  congruent  if 
two  sides  and  the  included  angle  of  one  are  equal  to  two 
sides  and  the  included  angle  of  the  other. 


Fig.  152 

Hypothesis:  In  [s] ABCD  and  A'B'CD\  w  =  w',  x  =  x\ 
and  ZA=  ZA'. 

Conclusion:     /Z7  ABCD  U  EJ  A 'B'CD'. 

Analysis:  To  prove  CJ  ABCD  ^  OJ  A'B'CD\  prove 
that  they  will  fit  when  superposed. 

Proof: 

STATEMENTS 

I.  Place  'eJABCD  upon  EJ  A'B'C'D'  so  that  the  sides 
of  ZA  will  fall  on  the  sides  of  Z.A\  x  lying  along  x' 
and  w  along  w' . 
II.  Point  B  will  fall  upon  point  B'. 

III.  Point  D  will  fall  upon  point  D'. 

IV.  1.    ZD   and    /.D'  are   supplements  of    ^  ^4  and  A\ 
respectively.      .'.  ZD=  AD'. 

2.  z  will  fall  along  the  line  of  z\ 

V 

Let  the  pupil  give  all  reasons  and  complete  the  proof. 


82  PLANE  GEOMETRY 

TESTS  FOR  PARALLELOGRAMS 

99.  The  definition  of  a  parallelogram  is  the  fundamental 
test  for  parallelograms.  Other  tests  are  dependent  primarily 
upon  the  fundamental  one. 

100.  Exercise.  Construct  a  quadrilateral  A  BCD  so  that  the 
opposite  sides  AD  and  BC  are  parallel  and  equal. 

Theorem  36.  If  a  quadrilateral  has  one  side  equal  and 
parallel  to  its  opposite,  it  is  a  parallelogram. 


Fig.   153 
Hypothesis:     In  the  quadrilateral  ABCD,  x  =  z  and  x\\z. 
Conclusion:     ABCD  is  a  ZZ7. 
Analysis  and  construction: 
I.  To  prove  ABCD  a  /Z7,  prove  w  \\  y. 
II.     *'       "      w\\y,  draw  AC  and  prove  Z  1  =  Z 2. 

III.     "       "      Z1=Z2,  prove 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 
101.  Exercise.     Construct  a  quadrilateral   ABCD  so  that  the 
opposite  sides,  AD  and  BC,  also  the  sides  AB  and  DC,  are  equal. 

Theorem  37.     If  a  quadrilateral  has  each  side  equal  to 
its  opposite,  it  is  a  parallelogram. 


Fig.   154 

Hypothesis:     In  the  quadrilateral  ABCD,  x  =  z  and  w  =  y. 

Conclusion:    ABCD  is  a  O. 

The  analysis  and  the  proof  are  left  to  the  pupil. 


QUADRILATERALS  83 

102.  Thp:orkm  3S.     If  the  diagonals   of  a   quadrilateral 
bisect  each  other,  the  quadrilateral  is  a  parallelogram. 

EXERCISES  INVOLVING  TESTS  FOR  PARALLELOGRAMS 

103.  To  prove  that  any  given  quadrilateral  is  a  parallelo- 
gram, prove  that  it  has 

(1)  Each  side  parallel  to  its  opposite,  or 

(2)  Each  side  equal  to  its  opposite,  or 

(3)  One  side  equal  and  parallel  to  its  opposite,  or 

(4)  The  diagonals  bisecting  each  other. 

1.  Construct  a  parallelogram,  given  two  sides  and  the  included 
angle.     In  how  many  ways  is  this  possible? 

2.  If  two  sides  of  a  quadrilateral  are  parallel  and  two  opposite 
angles  are  equal,  the  figure  is  a  parallelogram. 

3.  If  A  BCD  is  a  parallelogram  and  E  and  F  are  respectively 
the  mid-points  of  the  opposite  sides  AB  and 
CD,  prove  that  AECF  is  a  parallelogram. 

4.  Given  EJ  ABCD,  with  points  E  and  F 


on  the  diagonal  .4C  so  that  AE  =  CF,  prove  Fig.  155 

that  BFDE  is  a  parallelogram  (Fig.  155).  „ 

5.  Investigate  Ex.  4   if  points  E   and  F 
are  on  the  diagonal  extended. 


6.  Given  HJABCD,  AE  =  CG,  AH  =  CF, 
prove  that  EFGH  is  a  parallelogram  (Fig.  156).        Fig.  156 

7.  Construct  a  parallelogram,  given  the  sides  and  one  diagonal. 

THEOREMS  AND  EXERCISES  INVOLVING  TESTS  FOR 
EQUAL  AND  PARALLEL  SEGMENTS 

104.  Ths.  36  and  37,  together  with  the  definition  of  a  par- 
allelogram, give  an  additional  test  for  equal  and  for  parallel 

segments : 

To  prove  two  segments  equal  or  parallel,  find  a  quadri- 
lateral of  which  these  segments  are  opposite  sides  and  prove 
that  the  quadrilateral  is  a  parallelogram. 


84  PLANE  GEOMETRY 

Theorem  39.  Segments  of  parallels  intercepted  between 
parallel  lines  are  equal. 

Theorem  40.  Segments  of  perpendiculars  intercepted 
between  parallel  lines  are  equal. 

Ex.  1.  Construct  through  a  given  point  a  line  that  shall  be 
parallel  to  a  given  line. 

Ex.  2.  If  A  BCD  is  a  parallelogram  and  E  and  F  are  respec- 
tively the  mid-points  of  the  opposite  sides  A  B  and  CD,  prove  that 
AF\\CE. 

Ex.  3.     Given  CJABCD  with  E,  F,  G,  and  H  the  mid-points  of 
the  sides,  and  the  points  joined  as  indicated  in 
the  figure;  prove  that  AXCY  is  a  parallelogram 
(Fig.  157). 

Ex.  4.     In  what  other  way  may  the  points  ^          f^         b 
in  Fig.  157  be  joined  so  as  to  form  a  parallelo-  ftg.  157 

gram? 

Ex.  5.  Construct  a  line  that  shall  be  intercepted  by  the  sides 
of  a  given  triangle  and  be  parallel  to  the  base  of  the  triangle  and 
shall  have  a  given  length. 

SPECIAL  QUADRILATERALS 

105.  A  four-sided  figure  is  called  a  quadrilateral.  Unless 
otherwise  stated,  a  quadrilateral  should  be  drawn  with  no 
sides  equal  and  no  sides  parallel. 

The  following  special  kinds  of  quadrilaterals  are  important : 

A  quadrilateral  formed  by  two  isosceles  triangles  on  oppo- 
site sides  of  the  same  base  is  called  a  kite  (Fig.  158,  No.  1). 

A  quadrilateral  with  but  one  pair  of  parallel  sides  is  called 
a  trapezoid  (Fig.  158,  No.  2). 

If  a  trapezoid  has  its  non-parallel  sides  equal,  it  is  called 
an  isosceles  trapezoid  (Fig.  158,  No.  3). 

The  non-parallel  sides  of  a  trapezoid  are  sometimes  called 
the  legs  of  the  trapezoid. 

The  parallel  sides  of  a  trapezoid  are  called  the  bases. 


QUADRILATERALS 


85 


The  perpendicular  distance  between  the  bases  of  a  trape- 
zoid is  called  the  altitude  of  the  trapezoid.  In  Fig.  158, 
No.  2,  b  and  b'  are  the  bases  and  a  is  the  altitude. 

A  quadrilateral  with  two  pairs  of  parallel  sides  has  been 
defined  as  a  parallelogram. 

The  perpendicular  distance  between  the  bases  of  a  paral- 
lelogram is  called  the  altitude  of  the  parallelogram.  A  par- 
allelogram has  two  altitudes,  since  each  pair  of  parallel 
sides  may  be 'Considered  as  bases.  In  Fig.  158,  No.  4,  b 
may  be  considered  as  a  base  with  a  as  the '  corresponding 
altitude,  or  b'  may  be  considered  as  a  base  with  a'  as  the  cor- 
responding altitude. 


The  following  special  kinds  of  parallelograms  are  of  con- 
siderable importance  and  of  widespread  occurrence: 

A  parallelogram  with  one  right  angle  is  called  a  rectangle 
(Fig.  158,  No.  5). 

A  parallelogram  with  two  consecutive  sides  equal  is  called 
a  rhombus  (Fig.  158,  No.  6). 

A  rectangle  with  two  consecutive  sides  equal  is  called  a 
square  (Fig.  158,  No.  7). 

The  segment  joining  the  mid-points  of  two  opposite  sides 
of  a  quadrilateral  is  called  a  median  of  the  quadrilateral 
(segment  XY,  Fig.  158,  No.  8). 

The  definition  of  any  particular  figure  is  the  fimdamental 
test  for  the  determination  of  that  figure. 

7 


86  PLANE  GEOMETRY 

KITES 

106.  Ex.  1.     One  diagonal  of  a  kite  is  an  axis  of  symmetry. 

Ex.  2.  The  axis  of  symmetry  of  a  kite  bisects  the  angles 
through  whose  vertex  it  passes. 

Theorem  41.  The  diagonals  of  a  kite  are  perpendicular 
to  each  other,  and  the  one  which  is  an  axis  of  symmetry 
bisects  the  other. 

ISOSCELES  TRAPEZOIDS 

107.  Ex.  1.    The  base  angles  of  an  isosceles   trapezoid  are 

equal  and  the  diagonals  are  equal. 

Ex.  2.  The  segment  joining  the  mid-points  of  the  parallel 
sides  of  an  isosceles  trapezoid  is  an  axis  of  symmetry.  - 

Ex.  3.  If  the  base  angles  of  a  trapezoid  are  equal,  the  trape- 
zoid is  isosceles. 

Ex.  4.  Construct  an  isosceles  trapezoid,  given  the  two  bases 
and  the  altitude. 

Ex.  5.  Construct  an  isosceles  trapezoid,  given  two  consecutive 
sides  and  the  included  angle. 

RECTANGLES 

108.  Ex.  1.    Construct  a  rectangle,  given  two  adjacent  sides. 
Theorem  42.    All  the  angles  of  a  rectangle  are  right 

angles. 

Ex.  2.    What  properties  has  a  rectangle  by 
virtue  of  the  fact  that  it  is  a  parallelogram? 
Ex.  3.    The  diagonals  of  a  rectangle  are  equal. 


Ex.  4.     If  the  diagonals  of  a  parallelogram  are         ^^°-  ^^^ 
equal,  the  parallelogram  is  a  rectangle  (Fig.  159). 
Analysis: 

I.  To  ipToveOJABCD  a  n  ,  prove  ZA=aTt.Z. 

II.    •'       *'      Z^=art.  Z,  prove  Z.4  =  Z5. 

III.    "       •'      Z^  =  ZB,  prove 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 

Ex.  5.    Construct  a  rectangle,  given  one  side  and  one  diagonal. 


QUADRILATERALS  87 

Ex.  6.  Construct  a  rectangle,  given  one  diagonal  and  the 
angle  between  the  diagonals. 

Ex.  7.  The  medians  of  a  rectangle  bisect  each  other  at  right 
angles. 

Ex.  8.  Make  a  list  of  all  of  the  properties  of  the  rectangle. 
Which  of  these  properties  are  special  properties  of  the  rectangle? 

RHOMBUSES 

109.  Ex.  1.  Construct  a  rhombus,  given  one  side  and  one 
angle. 

Theorem  43.    All  the  sides  of  a  rhombus  are  equal. 

Theorem  44.  The  diagonals  of  a  rhombus  are  perpen- 
dicular to  each  other  and  bisect  the  angles  through  which 
they  pass. 

Ex.  2.    The  diagonals  of  a  rhombus  are  axes  of  symmetry. 

Ex.  3.  If  the  diagonals  of  a  quadrilateral  bisect  each  other 
at  right  angles,  the  figure  is  a  rhombus. 

Ex.  4.    The  altitudes  of  a  rhombus  are  equal. 

Ex.  5.     Construct  a  rhombus,  given  one  side  and  one  diagonal. 

Ex.  6.  Construct  a  rhombus,  given  one  side  and  the  altitude. 
When  is  this  problem  impossible? 

Ex.  7.  What  properties  has  a  rhombus  by  virtue  of  the  fact 
that  it  is  a  parallelogram?  Make  a  list  of  all  the  properties  of  the 
rhombus.     Which  of  these  are  special  properties  of  the  rhombus? 

SQUARES 

110.  Ex.  1.     Construct  a  square,  given  one  side. 

Ex.  2.  Show  that  a  square  may  be  classified  as  a  special  kind 
of  a  rectangle,  or  rhombus.  From  these  facts  make  a  list  of  all 
the  properties  of  the  square. 

Ex.  3.  Each  diagonal  and  each  median  of  a  square  is  an  axis 
of  symmetry. 

Ex.  4.  If  the  diagonals  of  a  quadrilateral  are  equal  and  bisect 
each  other  at  right  angles,  the  figure  is  a  square. 

Ex.  5.     Construct  a  square,  given  one  diagonal. 


88  PLANE  GEOMETRY 

PARALLELS   AND    SEGMENTS    ON    TRANSVERSALS 

TEST  FOR  EQUAL   SEGMENTS 

111,  Theorem  45.  If  a  series  of  parallels  cuts  off  equal 
segments  on  one  transversal,  it  cuts  off  equal  segments  on 
all  transversals. 


Fig.  160 

Hypothesis:  li,\\l2,\\h,\\U,  cut  the  transversal  h  so  that 
segments  A,  b,  and  c  are  equal,  and  cut  off  the  segments 
X,  y,  and  z  on  transversal  k. 

Conclusion :    x  =  y  =  z. 
Analysis: 
I.  To  prove  x  =  y  =  z,  draw  MN,  PQ,  and  RS  \\  h  and 
prove  AMNP  ^  APQR  ^  ARST. 

II.  To  prove  AMNP  m  APQR  \  ^^7=  Z2=  Z3 
mARST,  prove]  ^^^^g^^g'. 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 

Ex.  1.     Prove  Th.  45  by  drawing  the  construction  lines  from 
the  points  of  division  on  h  instead  of  from  those  on  k. 

Problem  7.  To  divide  a  given  segment  into  any  num- 
ber of  equal  parts. 

Solution:  Let  AB  he  the  given  segment.  From  point  A  draw  a 
ray  making  any  convenient  angle  with  AB.  From  point  A  lay  off 
equal  divisions  on  this  ray.  The  number  of  divisions  must  be  the 
same  as  the  number  of  parts  into  which  ^5  is  to  be  divided.  Join 
the  last  point  of  division  with  B.  Draw  parallels  from  the  other 
points  of  division.     Let  the  pupil  draw  the  figure  and  give  the  proof. 


QUADRILATERALS 

Ex.  2.  A  line  may  be  divided  into  any 
number  of  equal  parts  (for  example,  5)  by  the 
construction  shown  in  Fig.  161.  Give  the 
complete  directions  and  the  proof. 


Note.  A  sheet  of  ruled  paper  may  be  used 
to  divide  a  segment  into  a  given  number  of 
equal  parts.  Number  the  lines  ps  in  Fig.  162. 
If  the  segment  is  to  be  divided  into  7  equal 
parts,  put  the  ends  of  the  segment  on  lines  0 
and  7.    Why? 

Ex.  3.  Show  how  a  carpenter's  steel  square 
may  be  used  to  divide  a  board  into  strips 
of  equal  width  (Fig.  163). 

Note.    By  the  width  of  the  board  is  meant  the   ^ 
perpendicular  distance  between  the  sides.  Fig.   163 

RELATED   THEOREMS   CONCERNING   TRIANGLES 

112.  Theorem  46.    A  segment  parallel  to  the  base  of  a 
triangle  and  bisecting  one  side  is  equal  to  half  the  base. 


\  \  '\  \  . 

Jvr 
Fig.  161 

•'//' 

0       A 

t         \ 

2           \ 

:i           \ 

4              \ 

,5                 \ 

«                   \ 

7                           \ 

s                    Ji 

Fig.  162 

> 

/y\,:i 

<- 

% 

//     \\« 

^ 

I      ^         W.-/       ?* 

;^ 

N\i. 

K. 

Fig.   164 


Hypothesis:     In  AABC,  XY  bisects  AC  and  ||  AB. 

Conclusion:    XY  =  }/2  AB. 

Analysis  and  construction: 
I.  To  prove  XY  =  }/2  AB,  prove  XY  is  equal  to  a  seg- 
ment that  is  }/2  AB. 

II.   .-.   construct  YZ  from  Y\\AC  and  prove  XY=AZ 
Sind  XY  =  ZB. 

Let  the  pupil  give  the  proof. 


90  PLANE   GEOMETRY 

113.  Theorem  47.     A  segment  parallel  to  the  base  of  a 
triangle  and  bisecting  one  side  bisects  the  other  side  also. 

Suggestion.    Draw  a  line  through  the  vertex  of  the  triangle  parallel 
to  the  base  and  apply  Th.  45. 

114.  Theorem  48.     A  segment  bisecting  two  sides  of  a 
triangle  is  parallel  to  the  third  side. 


Fig.   165 

Hypothesis:     In  A  ABC,  XY  joins   X  and  Y,  the  mid- 
points oi  AC  and  BC  respectively. 

Conclusion:    XY\\AB. 
Analysis  and  construction: 
I.  To  prove  XY\\AB,  prove  that  XY  coincides  with  a 

segment  that  is  ||  to  AB. 
11.    /.  draw  XZ\\AB  from  X  and  prove  that  XZ  coin- 
cides with  XY, 
III.  To  prove  that  XZ  coincides  with  XY,  show  that  XZ 
and  XY  both  pass  through  X  and  Y. 

Proof : 

STATEMENTS 

I.  a.  XY  passes  through  X  and  Y. 

b.  XZ  passes  through  X. 

c.  XZ  passes  through  Y. 
11.  XY  and  XZ  coincide. 

III.  XYWAB, 
Let  the  pupil  give  reasons. 
Apply  Th.  47  in  Ic.     For  II  see  As.  6. 

Exercise.     A  segment  bisecting  two  sides  of  a  triangle  is  equal 
to  half  the  base. 


QUADRILATERALS  91 

115.  If  three  or  more  lines  pass  through  a  common  point, 
they  are  said  to  be  concurrent. 

Theorem  49.  The  medians  of  a  triangle  are  concurrent 
in  a  point  that  is  two-thirds  the  distance  from  each  vertex  to 
the  mid-point  of  the  opposite  side. 


Hypothesis:    ABC  is  any  triangle. 

Conclusion:  (1)  The  medians  are  concurrent.  (2)  The 
point  of  intersection  is  two-thirds  the  distance  from 
each  vertex  to  the  mid-point  of  the  opposite  side. 

Analysis  and  construction  for  (1) : 
I.  To  prove  that  the  medians  are  concurrent,  let  any 
two  medians,  as  AF  and  BG,  meet  at  O.     Join  CO 
and  extend  to  E  and  prove  that  AE  =  EB. 
II.  To  prove  that  AE  =  EB,  prove  that  AB  may  be  the 
diagonal  of  a  £17 . 

III.  .-.    extend  CE  to  H   so    that    OH  =  CO,    join   HB 

and  HA.    Prove  AH  BO  a  O. 

IV.  .-.  prove  HB  \\  OF  (part  of  AF)  and  AH  \\  OG  (part 
oiBG). 

The  proof  is  left  to  the  pupil. 

Analysis  for  (2) :     Prove  CO  =  %  CE. 

Exercise.      Segments    drawn    from    one 

vertex  of  a  parallelogram  to  the  mid-points        t/ 

of  the  opposite  sides   trisect   the    diagonal 

which  they  intersect  (Fig.  167). 

Fig.  167 

Suggestion.     Draw  diagonal  AC  and  apply  Th.  49. 


92  PLANE  GEOMETRY 

116.  Theorem  50.  The  median  from  the  vertex  of  the 
right  angle  of  a  right  triangle  to  the  hypotenuse  is  one-half 
the  h5rpotenuse. 


Hypothesis:  In  the  rt.  A  ABC,  AX  is  the  median  from 
the  /.A  to  the  hypotenuse  CB. 

Conclusion:    AX  =  }/2  CB. 

Analysis: 

I.  To  prove  that  AX  =  y2  CB,  prove  AX- X5  isosceles. 
II.     "        "  AX-XB  isosceles,  from  X  construct  XO  \\  AC 
and  prove  XO  a  ±  bisector  of  AB. 

Let  the  pupil  give  the  proof. 

117.  The  following  exercises  are  applications  of  Ths.  46-50. 

Ex.  1.  The  segments  joining  the  mid-points  of  the  sides  of 
a  triangle  divide  it  into  four  congruent  triangles. 

Ex.  2.     Construct  a  triangle,  given  the  mid-points  of  its  sides. 

Ex.  3.  Perpendiculars  from  the 'mid-points  of  two  sides  of  a 
triangle  to  the  third  side  are  equal.  How  might  this  exercise  be 
proved  if  the  given  triangle  were  isosceles? 

Ex.  4.  If  D  is  any  point  in  the  side  ^C  of  a  AABC,  the  seg- 
ments joining  the  mid-points  oi  AD,  DC,  CB,  and  AB  form  a 
parallelogram. 

Ex.  5.  Through  a  given  point  within  an  angle  draw  a  seg- 
ment terminated  by  the  sides  of  the  angle  and  bisected  by  the 
given  point. 

Ex.  6.     Divide  a  right  triangle  into  two  isosceles  triangles. 

Ex.  7.  Would  Ths.  46,  47,  and  48  be  true  for  parallelograms  as 
well  as  for  triangles?     Give  proofs. 


QUADRILATERALS  93 

TRAPEZOIDS 

118.  Theorem  51.  The  segment  joining  the  mid-points  of 
the  non-parallel  sides  of  a  trapezoid  is  parallel  to  the  bases. 

.L \ 

Fig.  169 

Hypothesis:  In  C^ABCD,  XY  joins  X  and  Y,  the  mid- 
points of  the  non-parallel  sides  AD  and  BC  respectively. 

Conclusion:    XY  \\  AB  and  CD. 
Analysis  and  construction: 
I.  To  prove  XY  \\  AB  and  therefore  ||  DC,  prove  that 

XY  coincides  with  a  segment  that  is  ||  AB. 
II..  .*.  draw  XZ  \\  AB  from  X  and  prove  that  XZ  coin- 
cides with  XY, 
III.  To  prove  that  XZ  coincides  with  XY,  show.   .    .    .  ■  . 
Let  the  pupil  complete  the  analysis  and  give  the  proof. 

119.  Theorem  52.  The  segment  joining  the  mid-points 
of  the  non-parallel  sides  of  a  trapezoid  is  equal  to  one-half 
the  sum  of  the  bases. 


~vO 


Fig.  170 

Analysis:    To  prove  A'F  =  K>  (AB+DC),  draw  D^  inter- 
secting XY  at  0  and  prove  XO  =  y^  AB  and  OY  =  }i  DC. 

Ex.  1.  Construct  a  trapezoid  so  that  AB-Q  cm.,  BC  =  3.2 
cm.,  CD=4:  cm.,  AD  =  2.S  cm. 

Ex.  2.  If  two  trapezoids  have  the  four  sides  of  one  equal 
respectively  to  the  four  sides  of  the  other,  the  angles  of  one  are 
equal  respectively  to  the  corresponding  angles  of  the  other. 


94 


PLANE  GEOMETRY 


SUPPLEMENTARY  EXERCISES 


EXERCISES  INVOLVING  PARALLELOGRAMS 

120.     Note.     Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend. 

1.  Make  a  review  diagram  for  Th.  36. 

2.  Any  segment  drawn  through  the  intersection  of  the  diago- 
nals of  a  parallelogram,  and  terminated  by  the  sides  of  the  paral- 
lelogram, is  bisected  by  the  point  of  intersection  of  the  diagonals. 

3.  Perpendiculars  drawn  to  a  diagonal  of  a  parallelogram 
from  the  opposite  vertices  are  equal. 

4.  The  bisectors  of  two  consecutive  angles  of  a  parallelogram  are 
perpendicular  to  each  other. 

5.  The  bisectors  of  two  opposite  angles  of  a  parallelogram 
are  parallel. 

6.  The  bisectors  of  the  angles  of  a  parallelogram  form  a 
rectangle. 

7.  The  median  to  one  pair  of  opposite  sides  of  a  parallelogram 
is  equal  and  parallel  to  the  other  two  sides. 

8.  The  medians  of  a  parallelogram  bisect  each  other. 

H 

9.  In  Fig.  171,  ABCD  is  a  parallelogram. 
The  sides  are  extended  through  A  and  C  so  that 
CE  =  AG  and  CF  =  AH.  Prove  that  EFGH  is 
a  parallelogram. 

10.  In  Fig.  172,  ABCD  is  a  parallelo- 
gram. CD  and  AB  are  extended  so  that 
DE  =  BF,  DF  cuts  CB  at  Y  and  BE  cuts 

DA  at  X.    Prove  DX  =  BY. 

Fig.  172 

11.  If,  in  Fig.  172,  DX  is  made  equal  to  BY,  and  BX  and 
DY  are  drawn  and  extended  to  meet  CD  at  E  and  AB  at  F 
respectively,  prove  DE  =  BF. 

12.  In  Fig.  173,  ABCD  is  a  parallelogram.  ^" 
DX  and  BY  are  perpendicular  to  ^C  from 
D    and    B    respectively.     DY  and   BX  are 
joined.     Prove  DXBY  a  parallelogram.  Fig.  173 


QUADRILATERALS 


95 


AE  =  CG    and 

D 


13.  In   Fig.  174,  A  BCD  is  a  parallelogram. 
AH  =  CF.     Prove  A£0^  ^  AGOF. 


14.  Investigate  the  case,  Ex.  13,  in  which    __^ 

E,  Hj  F,  and  G  are  on  the  side  of  the  paral-  pic.  174 

lelogram  extended.  From  a  Roman 


floor  design 

15.  In  Fig.  175,  OXZY  is  any  parallelogram, 
OC  is  any  line  through  O.  YA ,  XB,  and  ZC  are 
drawn  from  F,  X,  and  Z  respectively  perpendicu- 
lar to  OC.     Prove  OC  =  OA  +  OB. 

16.  Fig.  176  shows  two  forms  of  parallel  rulers 
often  used  in  mechanical  drawing  for  construct- 
ing parallel  lines.  Show  how  each  is  constructed 
and  upoii  what  theorems  in  geometry  the  con- 
struction depends.  Which  one  of  these  is  some- 
times used  in  folding  gates? 

17.  If  the  sides  of  ADEF  are  parallel  respec- 
tively to  the  sides  of  AABC  and  pass  through  the 
vertices  of  AABC,  prove  that  FE  =  2AB,  and  that 
BF  and  AC  bisect  each  other  (Fig.  177).  Fig.  177 

18.  If  the  opposite  angles  of  a  quadrilateral  are  equal,  the 
figure  is  a  parallelogram. 

19.  The  sum  of  the  perpendiculars  drawn  from  an  arbitrary 
point  in  the  base  of  an  isosceles  triangle  to  the  equal  sides  is  equal 
to  the  perpendicular  from  one  end  of  the  base  to  the  opposite  side. 

20.  Investigate  the  case,  Ex.  19,  in  which  the  arbitrary  point 
is  in  the  base  extended. 

21.  The  sum  of  the  perpendiculars  drawn  from  an  arbitrary 
point  within  an  equilateral  triangle  to  the  sides  is  equal  to  the 
altitude  of  the  triangle. 

22.  Investigate  the  case,  Ex.  21,  in  which  the  arbitrary  point 
is  outside  of  the  triangle. 


96  PLANE  GEOMETRY 

EXERCISES  INVOLVING  SPECIAL  QUADRILATERALS 

121.  1.  Are  the  diagonals  of  a  parallelogram  perpendicular  to 
each  other?  of  a  rhombus?  of  a  square?  of  a  trapezoid?  of  a 
kite?    of  a  rectangle?  , 

2.  Name  the  quadrilaterals  in  which  the  diagonals  are  equal 
to  each  other. 

3.  Name  the  quadrilaterals  in  which  the  angles  are  bisected 
by  the  diagonals. 

4.  What  quadrilaterals  have  one  axis  of  symmetry?  What 
quadrilaterals  have  two  axes  of  symmetry?  Have  any  quadri- 
laterals more  than  two  axes  of  symmetry?  Name  the  axis  or  the 
axes  in  each  case. 

5.  What  quadrilaterals  have  a  center  of  symmetry?  Name 
the  center  of  symmetry  in  each  case. 

6.  The  figure  formed  by  joining  the  mid-points  of  the  sides 
of  a  rectangle  is  a  rhombus. 

7.  The  figure  formed  by  joining  the  mid-points  of  the  sides  of 
a  rhombus  is  a  rectangle. 

8.  The  figure  formed  by  joining  the  mid-points  of  the  sides 
of  a  square  is  a  square. 

9.  In  Fig.  178,  A  BCD  is  a  square  with  the  equal 
distances  AE,  BF,  BG,  CH,  etc.,  measured  on  the 
sides  in  each  direction  from  the  vertices;  prove  that 
EHKN  and  FGLM  are  rectangles  and  that  XYZW    ^' 
is  a  square. 


^  £  JTB 

Fig.   178 


10.  Investigate  the  case,  Ex.  9,  in  which  A  BCD  is  a  rhombus 
or  a  rectangle  instead  of  a  square. 


11.  In  Fig.  179,  ABDC  is  a  rectangle,  AB  and 
CD  are  divided  into  the  same  number  of  equal 
parts  and  the  points  joined  as  indicated.  Prove 
that  the  figures  formed  are  rhombuses.  Would 
it  be  possible  to  construct  the  figure  so  that 
squares  are  formed  instead  of  rhombuses? 


C    K    L    M  y 


A     E    F    G    H 

Fig.  179 


QUADRILATERALS 


97 


12.  In  Fig.  180,  A  BCD  is  a 
square  with  its  diagonals  A  C  and  BD. 
AE  =  BF=CG  =  DH.  GW  and  YE 
are  parallel  to  AC,  and  FZ  and 
HX  are  parallel  to  DB.  Prove 
that  WXYZ  is  a  square. 

13.  Given  A  BCD  a  square  with 
i4 C  a  diagonal.  U  AX  =  CY,  prove 
that  5FZ)X  is  a  rhombus  (Fig. 
181).  Is  this  exercise  true  if 
A  BCD  is  a  rhombus?  If  A  BCD 
is  any  parallelogram? 


Fig.  180a 

Parquet 

floor 

design 


Fig.  181  Fig.  181a 

A  parquet  floor  design 


14.  Investigate  the  case,  Ex.   13,  hi  which  X  and  Y  are  on 
AC  extended. 

15.  In  Fig.  182  given  the  DABCD  with  £,  F,  G,  and  H  the 
mid-points  of  the    sides,    and   the    p_ 
points  joined  as  indicated.     Prove 
that  WXYZ  is  a  square. 

Analysis: 

To  prove  WXYZ  a  D,  prove 
it  a  a  with  WX  =  XY  and 
Zl=a  rt.  Z. 


Fig.  182  Fig.  182  a 

Parquet  floor  design 


■P    z 


16.  In  Fig.  183,  A  BCD  is  a  square  with  AX  =  BY 
CZ  =  DW.     Prove  that  PFXFZ  is  a  square.  ^ 


z   B 

Fig.  183 

From  a  Roman 

floor  design 


17.  In  Fig.  184,  ABC  is  an  isosceles  triangle 
with ZC  =  art.Z,C0±^5 from C.  AX=XY=YB. 
XW  and  YZ  are  ±  AB  from  X  and  Y.  Prove  that 
PFXFZ  is  a  square. 


Fig. 184 


18.  Construct  a  square  that  shall  have  its  vertices  on  the 
sides  of  a  rhombus. 


98  PLANE  GEOMETRY 

EXERCISES  INVOLVING  PARALLELS  AND  TRANSVERSALS 

122.     Note.     Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend. 

1.  Make  review  diagrams  for  Ths.  45,  49,  and  50. 

2.  Name  two  important  special  cases  of  Th.  45. 

3.  Prove  the  converse  to  each  of  the  theorems  called  for  in 
Ex.2. 

4.  The  segments  which  join  the  mid-points  of 
the  sides  of  a  quadrilateral  taken  in  order  form  a      '[/-/     ^f 
parallelogram  (Fig..  185). 

5.  Prove  Ex.  4   for  a  concave   quadrilateral     J^ ^ 

and  for  a  cross  quadrilateral.  Fig.  185 

6.  The  medians  of  a  quadrilateral  bisect  each  other. 

7.  The  segments  which  join  the  mid-points  of  two  opposite 
sides  of  a  quadrilateral  to  the  mid-points  of  the  diagonals  form  a 
parallelogram. 

8.  If  from  two  opposite  vertices  of  a  parallelogram  segments 
are  drawij  to  the  mid-points  of  the  opposite  sides,  these  segments 
trisect  the  diagonal  joining  the  other  two  vertices. 

9.  It  is  said  that  an  Arab,  about  900  a.d.,  trisected  a  given 
segment  as  in  Fig.  186.     Show  that  this  is  the  same 
construction  as  that  given  in  Ex.  8. 

10.  A  segment  joining  the  mid-points  of  the  non- 


parallel  sides  of  a  trapezoid  bisects  both  diagonals.      p^,^    jgg^ 

11.  The  segment  joining  the  mid-points  of  the  diagonals  of  a 
trapezoid  is  parallel  to  the  bases  and  equal  to  >^  their  difference. 

12.  If  one  leg  ot  a  trapezoid  is  perpendicular  to  the  bases,  the 
segments  joining  the  mid-point  of  the  other  leg  to  the  extremities 
of  the  first  leg  are  equal. 

13.  If  the  mid-points  of  the  legs  of  an  isosceles  triangle  are 
joined  to  the  mid-point  of  the  base,  the  figure  formed  is  a 
rhombus. 

14.  Given  A  BCD  a  trapezoid  with  the  base  AB  twice  the  base 
CD.  The  diagonals  AC  and  BD  intersect  at  0.  Prove  that  ^O  is 
twice  OC. 

Suggestion.     Join  the  mid-points  of  .40  and  OB  with  D  and  C. 


CHAPTER  V 

Inequalities 

ASSUMPTIONS    FOR   COMBINING   INEQUALITIES 

123.  To  As.  28  and  As.  29  in  chapter  iii,  which  were 
assumptions  of  inequality,  we  must  now  add  the  following: 

As.  31.  If  equal  segments  (or  angles)  are  added  to 
unequal  segments  (or  angles),  the  resulting  segments  (or 
angles)  are  unequal  in  the  same  order. 

As.  32.  If  equal  segments  (or  angles)  are  subtracted 
from  unequal  segments  (or  angles),  the  resulting  segments 
(or  angles)  are  unequal  in  the  same  order. 

As.  33.  If  unequal  segments  (or  angles)  are  added  to 
unequal  segments  (or  angles) ,  the  greater  to  the  greater  and 
the  lesser  to  the  lesser,  the  resulting  segments  (or  angles) 
are  unequal  in  the  same  order. 

As.  34.  If  unequal  segments  (or  angles)  are  subtracted 
from  equal  segments  (or  angles),  the  resulting  segments  (or 
angles)  are  unequal  in  the  opposite  order. 

As.  35.  If  unequal  segments  (or  angles)  are  mtdtiplied 
by  the  same  number,  the  resulting  segments  (or  angles)  are 
unequal  in  the  same  order. 

As.  36.  If  unequal  segments  (or  angles)  are  divided  by 
the  same  number,  the  resulting  segments  (or  angles)  are 
unequal  in  the  same  order. 

Exercise.  Illustrate  each  of  the  foregoing  assumptions,  using 
numbers  to  represent  the  lengths  of  the  segments  or  the  number 
of  degrees  in  the  angles. 

99 


100  PLANE  GEOMETRY 

FUNDAMENTAL   TESTS    OF   INEQUALITY 

124.  The  fundamental  test  of  inequality  is  As.  29,  given 
in  §57,  the  whole  is  greater  than  any  of  its  parts. 

TEST  FOR   UNEQUAL  ANGLES 

125.  The  fundamental  theorem  for  unequal  angles  is 
Th.  8,  an  exterior  angle  of  a  triangle  is  greater  than 
either  of  the  non-adjacent  interior  angles. 

TESTS  FOR  UNEQUAL   SEGMENTS 

126.  The  two  following  tests  for  unequal  segments  are 
important.  As.  37  may  be  called  the  fundamental  test  for 
unequal  segments.  It  is  known  to  every  one  who  goes 
'cross  lots  rather  than  around  the  corner. 

As.  37.  The  sum  of  two  sides  of  a  triangle  is  greater 
than  the  third. 

As.  38.  The  difference  between  two  sides  of  a  triangle 
is  less  than  the  third  side. 

Ex.  1.  How  many  triangles  can  be  formed  with  the  sides  7, 
11,  19,  and  15  cm.? 

Ex.  2.  If  two  sides  of  a  triangle  are  8  cm.  and  15  cm.,  what 
are  the  upper  and  lower  limits  of  the  third  side? 

Ex.  3.  If  D  is  an  arbitrary  point  in  the  side  BC  of  AABC, 
AB+BC+AC  >  2AD. 

Ex.  4.  The  sum  of  the  diagonals  of  any  quadrilateral  is  greater 
than  the  sum  of  either  pair  of  opposite  sides. 

Ex.  5.  The  perimeter  of*  a  quadrilateral  is  greater  than  the 
sum  of  its  diagonals. 

Ex.  6.  If  fiom  a  point  within  a  triangle  segments 
are  drawn  to  the  vertices  of  the  triangle,  the  sum 
of  these  segments  is  greater  than  >^  the  perimeter  of 
the  triangle.  /  / 

Ex.  7.     The  median  to  one  side  of  a  triangle  is  less       J'' 
than  }4  the  sum  of  the  other  two  sides  (Fig.  187).  Fig.  187 


INEQUALITVES."  101 

127.  Theorem  53.  If  from  a  point  within  a  triangle  seg- 
ments are  drawn  to  the  extremities  of  one  side,  their  sum 
is  less  than  the  sum  of  the  other  two  sides  of  the  triangle. 


Pig.  188 

Hypothesis:  Ift  AABC,  segments  DA  and  DB  are  drawn 
from  point  D  to  the  extremities  of  AB. 

Conclusion:    AD+DB  <  AC-\-CB. 

Analysis  and  construction:  To  prove  AD-\-DB<  AC-\-CB, 
extend  AD  to  meet  BC  at  E  and  prove  DB<  DE-\-EB  and 
AD-\-DE<AC+CE. 

Proof: 

STATEMENTS 

1.  DB<DE-\-EB. 

2.  AD+DE<AC+CE. 

3.  ,\  DB-\-AD-{-DE<DE+EB+AC+CE. 

4.  .\  DB+ADKEB+CE+AC. 

5.  :.  DB-\-AD<BC+AC. 
Let  the  pupil  give  the  reasons. 

For  3  use  As.  33.     For  4  use  As.  32. 

Ex.  1.  The  sum  of  the  distances  from  any  point  in  a  triangle  to 
the  vertices  is  less  than  the  perimeter  of  the  triangle.  • 

Ex.  2.     In  Fig.  188  prove  that  ZADB>ZC. 

Ex.  3.  If  O  is  any  point  in  the  side  AC  of  AABC,  OC+OB 
<AB+AC, 

8 


lOS  PLANE   GEOMETRY 

TESTS  FOR  UNEQUAL  SIDES  AND  ANGLES  IN  ONE 
TRIANGLE 
128.  Theorem  54.    If  one  angle  of  a  triangle  is  greater 
than  a  second,  the  side  opposite  the  first  angle  is  greater 
than  the  side  opposite  the  second  angle. 


Fig.   189 

Hypothesis:     In  AABC,  AB>  /.A. 
Conclusion:    AC  >  CB. 
Analysis  and  construction: 
I.  To  prove  AC  >  CB,  construct  a  triangle  with  BC  for 

one  side  such  that  the  sum  of  the  other  two  sides 

shall  equal  AC. 
II.    /.  construct  OB  from  point  B  so  that  OB  =  OA,  that 

is,  so  that  Z  2  =  Z  1,  and  compare  AC  with  OC-^OB 

and  OC+OB  with  CB. 

The  proof  is  left  to  the  pupil. 

129.  Theorem  55.  If  one  side  of  a  triangle  is  greater  than 
a  second,  the  angle  opposite  the  greater  side  is  greater  than 
the  angle  opposite  the  lesser  side. 


Fig.  190 

Hypothesis:     In  AABC,  AOCB. 
Conclusion :     AB>  /.A. 


INEQUALITIES  103 

Analysis  and  construction: 

I.  To  prove  ZB>ZA,  compare    ZB  with   an   angle 
that  is  greater  than  /.A. 
II.   .*.  on  AC  take  CD  =  CB,  draw  BD,  and  compare  Z  1 
with  ZB  and  ZA. 
Proof: 

^      STATEMENTS 

I.  a.   Zl=  Z2. 

b.  ZB  >Z2. 

c.  :.ZB>Z\. 
11.    Zl>ZA. 

III.   .-.  ZB>  ZA. 
Let  the  pupil  give  the  reasons. 

Ex.  1.  Review  Ths.  3  and  24  concerning  the  equal  sides  and 
angles  of  a  triangle. 

Ex.  2.     Prove  Th.  55  by  an  indirect  proof. 

Ex.  3.  Prove  Th.  55  by  the  following  construction:  '  Bisect 
Z  C,  continue  the  bisector  to  meet  ABaX  E.  On  CA  take  CD  =  CB. 
Join  E  and  D. 

Ex.  4.  In  Fig.  191,  XO  is  the  perpen- 
dicular bisector  of  AB.  ZW  \\  AB.  If  point 
C  moves  to  the  right  and  to  the  left  along  ZW, 
how  will  the  relative  lengths  of  CA  and  CB 
change?    What  will  be  the  limits  of  Z CAB  ? 

How  will  the  relative  sizes  ol    AA   and  B 
.  ^  Fig.  19i 

change? 

Ex.  5.  The  angles  at  the  extremities  of  the  greater  side  of  a 
triangle  are  acute. 

Ex.  6.     If,  in  a  ^  BCD,  BC  <  CD,  the  diagonal 
DB  does  not  bisect  Z  Z),  but  Z  2  <  Z  1  (Fig.  192). 

Ex.  7.     State  and  investigate  the  converse  of  ^^ 
Ex.  6.     Prove  your  conclusion.  Fig.  192 

Ex.  8.  An  angle  of  a  triangle  is  right,  acute,  or  obtuse  accord- 
ing as  the  median  from  its  vertex  is  equal  to,  greater  than,  or  less 
than  half  the  side  that  it  bisects. 


104      .  PLANE  GEOMETRY 

DISTANCES  AND   OBLIQUE   SEGMENTS 
130.  Theorem  56.    The  perpendicular    is   the    shortest 
segment  from  a  point  to  a  straight  line. 

The  proof  is  left  to  the  pupil. 

Ex.  1.  The  altitude  to  one  side  of  a  triangle  is  less  than  half 
the  sum  of  the  other  two  sides. 

Ex.  2.  The  sum  of  the  three  altitudes  of  a  triangle  is  less  than 
the  perimeter. 

The  distance  between  a  point  and  a  line  is  defined  as  the 
length  of  the  perpendicular  from  the  point  to  the  line. 

Theorem  57.  If  from  a  point  in  a  perpendicular  to  a 
straight  line  two  oblique  segments  are  drawn  cutting  the 
straight  line  at  unequal  distances  from  the  foot  of  the  per- 
pendicular, the  more  remote  is  the  greater. 


0'  o 

Fig.   193 


Hypothesis:     AO  _L  line  /  and  the  oblique  segments  AB 
and  AC  are  so  drawn  that  OB>OC. 
Conclusion:    AB>AC. 
Analysis: 
I.  To  prove  AB  >  AC,  prove  Z  2  >  Z  1. 
II.    *•       "      Z2>Z1,    "      Z 2  an  obtuse  angle. 
III.    "       "      Z2  an  obtuse  angle,  prove  Z2  >  Z4. 

Theorem  58.  If  from  a  point  in  a  perpendicular  to  a 
straight  line  two  unequal  oblique  segments  are  drawn,  the 
greater  cuts  the  straight  line  at  the  greater  distance  from 
the  foot  of  the  perpendicular. 

Suggestion.    Use  an  indirect  proof. 


INEQUALITIES  105 

TESTS  FOR  UNEQUAL  SIDES  AND  ANGLES  IN  TWO 
TRIANGLES 
131.  Theorem  59.  If  two  triangles  have  two  sides  of  one 
equal  to  two  sides  of  the  other,  but  the  included  angle  of  one 
greater  than  the  included  angle  of  the  other,  the  third  side 
of  the  first  is  greater  than  the  third  side  of  the  second. 


Fig.  194 

Hypothesis:  In  ^ABC sxid  A' B'C\AB  =  A' B\BC  =  B'C\ 
and  ZB>  ZB\ 

Conclusion:    AC>A'C. 

Analysis  and  construction: 

I.  To  prove  AC>A'C\  compare  AC  with  a  segment 
that  is  greater  than  A'C 

II.  .-.  place  AA'B'C  on  £\ABC  so  that  A'B'  coin- 
cides with  AB,  A'  on  A,  and  B'  on  B.  Bisect 
LC'BC\  let  the  bisector  meet  AC  at  X\  join  XC . 
Compare  AXA-XC  with  AC  and  AC, 

III.  To  compare  AX-\-XC'  with  AC,  prove  XC  =  XC. 
Let  the  pupil  complete  the  analysis  and  give  the  proof.  i 

Discussion.  In  constructing  Fig.  194  according  to  state- 
ment II  of  the  analysis  it  may  happen  that  point  C  will  fall 
on  the  Hne  AC  or  within  the  triangle  as  well  as  in  the  posi- 
tion shown  in  the  figure,  li  C  falls  within  the  triangle,  the 
proof  is  the  same  as  for  the  case  given  above.  If  C  falls 
on  AC,  the  theorem  is  evident  without  proof. 

Exercise.    If,  in  EJABCD,   ZA<ZB,  the  diagonal  opposite 
Z  i4  is  less  than  the  diagonal  opposite  Z  B. 


106 


PLANE  GEOMETRY 


132.  Theorem  60.  If  two  triangles  have  two  sides  of 
one  equal  to  two  sides  of  the  other,  but  the  third  side  of 
one  greater  than  the  third  side  of  the  other,  the  angle 
opposite  the  third  side  of  the  first  is  greater  than  the 
angle  opposite  the  third  side  of  the  second. 

Suggestion.    Use  an  indirect  proof. 

Exercise.  Review  any  theorem  or  theorems  concerning  congru- 
ent triangles  that  are  closely  related  to  Ths.  59  and  60. 


SUPPLEMENTARY  EXERCISES 

133.  Note.     Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend. 

1.  Name  the  theorems  that  are  tests  for  unequal  angles;  for 
unequal  segments. 

2.  If,   in   AABC,  the  median  AD  is  drawn  and   ZADB  is 
acute,  prove  AC>  AB. 

3.  In  quadrilateral  A  BCD,  AD=BC  and  ZD> 
jLC,    Prove  LB  >  ZA. 

4.  CD  is  the  median  to  side  AB  in  AABC,  and 
ZB  >  ZA.  If  any  point  X  in  CD  is  joined  to  A 
and  B,  prove  ZXBA  >  ZXAB. 

5.  In  Fig.  195  prove  that  ^X+^^<^I^+^I^. 
^^'  is  bisected  at  right  angles  by  ZY.  J5^'  is  a 
straight  line.  Y  is  any  point  in  ZY.  Prove  also 
that  Zr=Z2. 


Pig.  196 


Note.  Ex.  5  illustrates  some  important  facts  from  physics.  In 
Fig.  196,  XY  represents  a  plane  mirror  with  a  candle  C  in  front  of  it. 
E  represents  the  eye.  The  reflection  C  of  the 
candle  seems  to  be  as  far  behind  the  mirror  as  C  is 
in  front  of  it.  The  light  from  C  strikes  the  mirror  at 
M  and  is  turned  back  to  the  eye  at  E  so  that  the 
distance  CME  is  the  least  possible.  The  light 
appears  to  come  directly  from  C.  Also  Zl=  Z2. 
Similar  relations  hold  when  an  elastic  object  strikes 
a  surface  and  rebounds  freely. 


»c' 


Fig.  1! 


INEQUALITIES 


107 


in 


■fB 


6.  Show  how  to  find  the  path  of  a  billiard  ball  which 
is  at  A  (Fig.  197)  and  which  is  struck  so  as  to  rebound 
from  the  side  of  the  table  XY  and  strike  ball  B. 

Fig.  197 

7.  Show  the  path  of  a  billiard  ball  which  is  struck  so  as  to 
rebound  from  each  side  of  the  table  and  return  to  its  original 
position.  (From  O.  Henrici,  Elementary  Geometry,  Congruent 
Figures.) 

8.  Prove  Th.  54  by  the  construction  shown  in 
Fig.  198. 

A  nalysis: 
I.  To  prove  i4C>CS,  construct  a  triangle  with  AC 

for  one  side  such  that  the  difference  between 

the  other  two  sides  will  be  BC. 
II.    .'.  construct    Z1=Z2  and  compare 

9.  Prove   Th.   55  by  the   construction   shown 
Fig.  199. 

Analysis: 
I.  To  prove  Z4  >  Zl,  compare  Z4  with  an  angle  that 
is  greater  than  Zl. 
II.    .'.  construct  ^O  so  that  Z2=  Z3  and  compare  .  . 

10.  If  two  opposite  sides  of  a  quadrilateral  are  equal  but  the 
diagonals  are  unequal,  the  angles  which  are  opposite  the  longer 
diagonal  are  respectively  greater  than  the  angles  which  are  opposite 
the  shorter  diagonal. 

11.  If  two  sides  of  a  triangle  are  unequal,  the  median  drawn 
to  the  shorter  side  is  longer  than  the  median  drawn  to  the  longer 
side  (Fig.  200). 

Analysis: 

I.  To  prove  CX  >  AY,  prove  CO  >  AO. 
(See  Th.  49.) 

II.  To  prove  CO  >  AO,  draw  the  median  BZ  and 
compare  ACZO  and  AAZO. 

III.  To  compare   ACZO  and  AAZO,  prove  Zl>' 

Z2. 

IV.  To  prove  Zl  >Z2,  compare  ACZB  and  AAZB, 


CHAPTER  VI 
Circles  and  Related  Lines 

INTRODUCTORY 

DEFINITIONS 

134.  We  have  already  defined  circle,  radius,  diameter, 
chord,  and  arc.     (See  §12.) 

Two  circles  or  two  arcs  that  can  be  made  to  coincide  are 
called  congruent  circles  or  arcs. 

In  succeeding  chapters  we  shall  consider  two  ways  of 
measuring  an  arc,  namely:  by  its  length  and  by  the 
number  of  degrees  that  it  contains.  Each  method  gives  a 
numerical  measure  for  the  arc,  but  the  measures  and  the 
methods  are  different.  Before  two  arcs  can  be  made  to 
coincide  they  must  have  not  only  the  same  measure  but  the 
same  radius.  Congruent  arcs  will  have,  the  same  measure 
whichever  method  is  used  in  measuring  them  and  will  be 
called  equal  arcs. 

The  chord  joining  the  ends  of  an  arc  is  called  the  chord  of 
the  arc.  Every  arc  has  one  and  only  one  chord.  Every 
chord,  however,  has  two  arcs.  If  the  chord  is  a  diameter, 
its  two  arcs  are  congruent  and  are  called  semicircles.  If 
the  chord  is  not  a  diameter,  its  two  arcs  are  unequal.  The 
larger  arc  of  a  chord  is  called  the  major  arc  and  the  smaller 
arc  is  called  the  minor  arc. 

An  angle  with  its  vertex  at  the  center  of  a  circle  is  called 
a  central  angle.  The  sides  of  the  angle  cut  off  two  arcs  on 
a  circle.  The  minor  arc  cut  off  by  the  sides  of  a  central 
angle  is  said  to  be  the  arc  intercepted  by  the  angle, 

108 


CIRCLES  AND   RELATED   LINES  109 

ASSUMPTIONS    CONCERNING    CIRCLES 

136.  In  §29  we  have: 

As.  9.     Circles  with  equal  radii  are  congruent. 

As.  10.     Congruent  circles  have  equal  radii. 

To  these  two  assumptions  we  shall  add  the  following: 

As.  39.     The  diameter  of  a  circle  is  twice  its  radius. 

As.  40.  A  circle  is  located  definitely  if  its  center  and 
its  radius  are  known. 

As.  41.  If  a  line  passes  through  a  point  within  a  circle, 
the  Hne  and  the  circle  intersect  in  two  and  only  two  points. 

As.  42.     Every  diameter  bisects  the  circle. 

As.  43.  A  circle  is  symmetric  with  respect  to  any  diame- 
ter as  an  axis  and  with  respect  to  its  center  as  a  center. 

As.  44.  Between  the  same  two  points  on  a  circle  there 
is  one  and  only  one  minor  arc  of  the  circle,  provided  these 
points  are  not  the  ends  of  a  diameter. 

As.  45.  A  segment  joining  a  point  within  a  circle  and 
the  center  is  shorter  than  the  radius. 

As.  46.  If  a  segment  that  has  one  end  at  the  center  of  a 
circle  is  shorter  than  the  radius,  it  lies  wholly  within  the 
circle. 

As.  47.  A  segment  joining  a  point  without  a  circle  and 
the  center  is  longer  than  the  radius. 

As.  48.  If  a  segment  that  has  one  end  at  the  center  of 
a  circle  is  longer  than  the  radius,  it  extends  without  the 
circle  and  cuts  the  circle  but  once. 

As.  49.  In  the  same  circle  or  in  congiiient  circles  equal 
central  angles  intercept  equal  minor  arcs. 

As.  50.  In  the  same  circle  or  in  congruent  circles  equal 
minor  arcs  intercept  equal  central  angles. 

Note.  Ass.  49  and  50  should  be  verified.  Draw  the  figures  on 
fairfy  thin  paper,  place  the  centers  together,  and  hold  to  the  light. 
Figures  may  be  drawn  to  illustrate  Ass.  39-48. 

The  assumptions  in  §  30  are  true  fpr  ^rcs  of  the  same  circle. 


110  PLANE  GEOMETRY 

RELATED  ARCS,  CHORDS,  AND  CENTRAL  ANGLES 

136.  Ass.  49  and  50  are  closely  related  to  the  two  follow- 
ing theorems  and  should  be  learned  with  them. 

Theorem  61.     In  the  same  circle  or  in  congruent  circles 

A.  Equal  chords  intercept  equal  central  angles. 

B.  Equal  central  angles  intercept  equal  chords. 


Fig.  201 
Suggestion.    Prove  by  congruent  triangles. 
Theorem  62.    In  the  same  circle  or  in  congruent  circles 

A.  Equal  chords  have  equal  minor  arcs. 

B.  Equal  minor  arcs  have  equal  chords. 
Analysts  A: 

To  prove  BC  =  ZT,  prove  /.A=  ZX. 
Use  Th.  61A  and  As.  49. 

Analysis  B: 

To  prove  BC  =  ZY,  prove  /.A=  AX. 
Use  As.  50  and  Th.  61 B. 

Ex.  1.  If  two  circles  are  not  congruent,  can  an  arc  of  one  be 
congruent  to  an  arc  of  the  other? 

Ex.  2.  For  what  special  cases  do  the  proofs  of  Ths.  61  and 
62  have  no  meaning? 

Ex.  3.     Show  how  to  bisect  a  given  arc.  « 

Ex.,  4.    A  ray  from  the  center  of  a  circle  through  the  mid-point 

of  a  chord  is  perpendicular  to  the  chord  and  bisects  the  arc  of  the 

chord. 


CIRCLES  AND  RELATED  LINES  111 

CHORDS  IN  GENERAL 

FUNDAMENTAL   THEOREM 

137.  Theorem  63.     A  radius  perpendicular  to  a  chord 
bisects  the  chord  and  its  arc. 


•     Fig.  202 

Hypothesis:    The  OO  is  any  circle;  AB  is  any  chord  of 
OO,  and  the  radius  OC  ±  AB. 
Conclusion:    CO  bisects  the  AB  and  the  AB. 
The  analysis  and  the  proof  are  left  to  the  pupil. 

Ex.  1.  Investigate  the  case  in  which  AB  (Fig.  202)  is  a 
diameter. 

Ex.  2.  Construct  through  a  given  point  within  a  circle  a  chord 
that  shall  be  bisected  at  the  given  point. 

Ex.  3.  If  a  diameter  is  perpendicular  to  a  chord,  the  quadri- 
lateral formed  by  joining  the  extremities  of  the  chord  to  the 
extremities  of  the  diameter  is  a  kite. 

Ex.  4.  If  two  circles  have  the  same  center  and  a  line  intersects 
both  of  them,  the  segments  intercepted  between  the  circles  are 
equal. 

Suggestion.  Draw  a  radius  perpendicular  to  the  given  line  and  sub- 
tract equal  segments  from  equal  segments. 

Ex.  5.  A  segment  from  the  center  of  a  circle  to  the  mid-point 
of  an  arc  is  a  perpendicular  bisector  of  the  chord  of  the  arc. 

Ex.  6.  If,  in  a  circle  whose  center  is  0,  B  is  the  mid-point  of 
ACf  perpendiculars  from  B  to  AO  and  CO  are  equal. 

Ex.  7.  If  two  circles  intersect,  the  segment  that  joins  the 
centers  bisects  the  common  chord  at  right  angles. 

Suggestion^     Show  that  a  kite  is  formed  by  the  radii  drawn  to  the 
points  of  intersection. 


112 


PLANE  GEOMETRY 


TEST  FOR  DIAMETERS 
138.  Theorem  64.    The  perpendicular  bisector  of  a  chord 
passes  through  the  center  of  the  circle. 


11. 


IIL 


Fig.  203 
Hypothesis:     OO  is  any  circle;  AB  is  any  chord  of  OO; 
Xy  is  the  _L  bisector  of  AB. 

Conclusion:    XY  passes  through  point  O. 
Analysis  and  construction: 
I.  To  prove  that  XY  passes  through  point  0,  prove  that 
XY  coincides  with  a  line  that  passes  through  O. 
.*.   construct  OZ  from  0  J_  AB  and  prove  that  XY 

and  OZ  coincide. 
To  prove  that  XY  and  OZ  coincide,  show  that  they 
are  both  ±  bisectors  of  AB. 
Proof: 

STATEMENTS 

I.  a.  Xy  is  the  _L  bisector  of  A6. 
6.  OZ  _L  AB. 
c.  OZ  bisects  AB. 
II.  XY  coincides  with  OZ. 
III.  XY  passes  through  point  O. 

CONSTRUCTION  AND  DEFINITE  LOCATION  OF  CIRCLES 
139.  Ex.  1.     To  find  the  center  for  a  given  arc. 
Suggestion.     Find  the  intersection  of  two  diameters. 

Ex.  2.     To  construct  a  circle  that  shall  pass  through  the  three 
vertices  of  a  triangle. 

Suggestion.    The  sides  of  the  triangle  will  be  chords  of  the  circle. 


REASONS 

I.  a.  Given. 

h.  Construction. 
c.  Why? 

II.  (See  As.  7.) 
IIL  Why? 


CIRCLES  AND   RELATED   LINES  113 

Theorem  65.  One  and  only  one  circle  can  be  drawn 
through  three  non-collinear  points. 

The  proof  is  left  to  the  pupil. 

We  have,  therefore,  two  methods  of  locating  circles 
definitely : 

a.  If  the  center  and  radius  are  known,  the  circle  is 
located  definitely  (As.  40). 

6.  If  a  circle  passes  through  three  given  non-collinear 
points,  it  is  located  definitely  (Th.  65). 

TEST   FOR   EQUAL   CHORDS 

140.  Theorem  66.  If  in  the  same  circle  or  in  congruent 
circles  perpendiculars  from  the  center  to  two  chords  are 
equal,  the  chords  are  equal. 


Fig.  204 

Hypothesis:     OA  =  eX,  AD  ±BC  from  A,  XZ  ±YW 
iromX,  SindAD  =  XZ. 


Conclusion :    BC=YW, 

Analysis  and  construction:    To  prove  BC—YW,  prove 

CD=wz,  CD=y2  CB,  wz=y2  yw. 

Let  the  pupil  complete  the  analysis  and  give  the  prooL 
Exercise.     State  all  tests  for  equal  chords. 

EQUAL   DISTANCES 

141.  Theorem  67.  In  the  same  circle  or  in  congruent 
circles  perpendiculars  from  the  center  to  two  equal  chords 
are  equal.     (See  Fig.  204.) 

Let  the  pupil  give  the  analysis  and  the  proof. 


114 


PLANE   GEOMETRY 


TANGENTS 
TESTS   FOR  TANGENTS 

142.  A  line  that  touches  a  circle  at  one  point  but  does  not 
cut  it  is  called  a  tangent  to  the  circle.  This  definition  is  the 
fundamental  test  for  tangents. 

The  point  at  which  the  tangent  touches  the  circle  is  called 
the  point  of  contact  or  the  point  of  tangency  of  the  tangent. 

143.  Theorem  68.  A  line  which  is  perpendicular  to  a 
radius  at  its  outer  extremity  is  a  tangent  to  the  circle. 


Fig.  205 

Hypothesis:     OO  is  any  circle;  radius  OA  is  any  radius; 
line  AB  ±  OA  at  A. 

Conclusion:     AB  is  tangent  to  OO  at  A. 

Analysis  and  construction: 

i.  To  prove  that  AB  is  tangent  to  OO  at  A,  prove  that 

all  points  in  AB  except  A  lie  outside  the  circle. 
II.  To  prove  that  any  point  in  AB  other  than  A,  such  as 
-  M,  lies  outside  the  circle,  join  O  and  M  and  prove 

OM  >  OA. 
Proof: 

STATEMENTS 
I.    OM    >   OA. 

II.    .'.  M  lies  outside  OO. 

III.  All  points    except    A    lie 
outside  OO. 

IV.    .-.  AB  is  tangent  to  OO.      IV.  Whyi 


REASONS 

I.  Th.  56  (quote  in  full). 

II.  As.  48. 

III.  Since  M  is  any  point  in 
AB  other  than  A. 


CIRCLES  AND  RELATED  LINES  115 

144.  Problem  8.  To  construct  a  tangent  to  a  circle  at 
a  given  point  on  the  circle. 

Ex.  1.  Construct  a  circle  of  given  radius  tangent  to  a  given 
line  at  a  given  point. 

Ex.  2.  Show  that  only  one  tangent  can  be  drawn  to  a  circle 
at  a  given  point  on  the  circle. 

TEST  FOR  PERPENDICULARS 

145.  Theorem  69.  A  tangent  to  a  circle  is  perpendicular 
to  the  radius  drawn  to  the  point  of  contact. 


Fig.  206 

Hypothesis:     OO  is  any  circle,  AB  is  tangent  to  OO  at 
A .     OA  is  the  radius  drawn  to  the  point  of  contact  A . 
Conclusion:    AO  ±  AB. 
Analysis  and  construction: 
L'  To  prove  AO  ±  AB,  show  that  the  supposition  that 

AB  is  not  J_  AO'^leads  to  an  absurdity. 
II.  If  AO  is  not  J_  AB,  suppose  some  other  line,  as  DO^ 
is  ±  AB,  and  show  that  the  supposition- that  DO  is 
±  AB  contradicts  the  hypothesis. 
Ex.  1.     Two  tangents  at  the  ends  of  a  diameter  are  parallel. 
Ex.  2.     A  diameter  bisects  all  chords  that  are  parallel  to  the 
tangents  at  its  extremities. 

LENGTHS  OF  TANGENTS;   TEST  FOR  EQUAL  SEGMENTS 
IIG.  Theorem  70.     If  two  tangents  meet  at  a  point  with- 
out a  circle,  the  distances  from  the  intersection  to  the  points 
of  tangency  are  equal. 

The  analysis  and  the  proof  are  left  to  the  pupil. 


116  PLANE  GEOMETRY 

TEST  FOR  DIAMETERS 

147.  Theorem  71.    A  perpendicular  to  a  tangent  at  the 
point  of  contact  passes  through  the  center  of  the  circle. 


Fig.  207 

Hypothesis:     In  QO,  AB  is  tangent  to    OO  at  A   and 
AC  ±  AB  Sit  A. 

Conclusion:    AC  passes  through  0. 

Analysis  and  construction: 

I.  To  prove  that  AC  passes  through  0,  prove  that  AC 

coincides  with  a  Hne  that  does  pass  through  0. 
11.    .*.    connect  0  and  A  and  prove  that  AO  and  AC 

coincide. 
III.  To  prove  that  AO  and  AC  coincide,  show  that  they 

are  both  ±  to  AB  at  A.     (See  As.  7.) 
The  proof  is  left  to  the  pupil. 

EXERCISES  INVOLVING  TANGENTS 

148.  1.  Construct  a  line  that  shall  be  tangent  to  a  given  circle 
and  parallel  to  a  given  line. 

2.  Construct   a  line   that   shall  be  tangent  to  a  given  circle 
and  perpendicular  to  a  given  line. 

3.  Construct  a  line  that  shall  be  tangent  to  a  given  circle  and 

make  a  given  angle  with  a  given  line. 

* 

4.  A  tangent  to  a  circle  at  the  mid-point  of  an  arc  is  parallel 
to  the  chord  of  the  arc. 

5.  If  two  circles  have  the  same  center,  a  chord  of  the  larger 
which  is  a  tangent  of  the  inner  is  b^isected  at  the  point  of  tangency. 


CIRCLES  AND  RELATED  LINES  117 

6.  If  two  circles  have  the  same  center,  chords  of  the  larger  which 
are  tangents  of  the  inner  are  equal. 

7.  If  two  tangents  to  a  circle  meet  at  a  point 
without  the  circle,  the  bisector  of  the  angle  between 
the  tangents  passes  through  the  center  of  the  circle. 

8.  Fig.  208  shows  an  instrument  that  may  be 
used  to  find  the  center  of  a  metal  disk.  How  is  it 
made  and  how  would  it  be  used? 

TWO  CIRCLES  AND  RELATED  LINES 

DEFINITIONS 

@©6)999 

No.  1  So.  2  yo.  3  So.  4  So.  5  No.  6 

Fig.  209 

149.  Fig.  209  shows  the  six  possible  relations  of  two 
circles. 

Two  circles  are  said  to  be  concentric  if  they  have  the 
same  center  (Fig.  209,  No.  1). 

Two  circles  are  said  to  be  tangent  if  they  have  but  one 
common  point.  They  may  be  tangent  internally  as  in 
Fig.  209,  No.  3,  or  tangent  externally  as  in  Fig.  209,  No.  5. 
This  definition  is  the  fundamental  test  for  tangent  circles. 

The  line  passing  through  the  centers  of  two  circles  is 
called  the  line  of  centers  of  the  two  circles. 

The  segment  through  the  points  of  intersection  of  two 
circles  is  called  the  common  chord  of  the  two  circles. 

150.  Since  any  diameter  of  a  circle  is  an  axis  of  symmetry 
of  that  circle,  we  will  assume 

As.  51.  Th^  line  of  centers  of  two  circles  is  an  axis  of 
sjnnmetry  of  the  two  circles. 

Exercise.    When  have  two  circles  a  center  of  symmetry? 
9 


lis  .  PLANE  GE:OMETRY 

INTERSECTING   CIRCLES 

151.  Theorem  72.     If  two  circles  intersect  in  one  point 
not  on  the  line  of  centers,  they  intersect  in  two  points. 


Fig,  210 

Hypothesis:     (DA  and  B  intersect  at  point  P  not  on  the 
line  of  centers  AB. 

Conclusion:     (DA  and  B  intersect  in  a  second  point. 

Analysis:  To  prove  that  (DA  and  B  intersect  at  a 
second  point,  prove  that  there  is  on  each  circle  one 
point  P'  which  is  symmetric  to  point  P. 

Cor.    If  two  circles  intersect,  the  points  of  intersection 
are  symmetric  points. 

We  will  assume  as  evident  that 

As.  52.     Two  circles  cannot  intersect  at  more  than  two 
points. 

Theorem  73.     If  any  two  circles  intersect,  the  line  of 
centers  is  the  perpendicular  bisector  of  the  common  chord. 

Suggestion.     Prove  by  folding  the  figure  on  the  axis  of  symmetry. 

Ex.  1.     Use  Th.  73  to  construct  a  ±  bisector  to  a  given  seg- 
ment and  to  construct  a  _L  to  a  line  from  a  point  not  in  the  line. 

Ex.  2.     If  two  equal  circles  intersect,  the  common  chord  sub- 
tends equal  central  angles  in  the  two  circles. 

Ex.  3.     In  Fig.   211,    ®0   and   0'  are  two 
equal  (D  intersecting  at  A  and  B.     The  line  of    cf 
centers  meets   the    ©0   at    C   and  QO'  at  D. 
Prove  ZACB=ZADB, 


CIRCLES  AND   RELATED   LINES 


119 


152.  Theorem  74.     If   two   congruent   circles   intersect, 
the  common  chord  is  an  axis  of  symmetry  of  the  figure. 


Fig.  212 

Suggestion.  Fold  the  figure  on  PP'  as  an  axis.  Point  A  will  fall 
on  point  B.     The  circles  will  coincide. 

Cor.  If  two  congruent  circles  intersect,  the  segment 
joining  the  centers  and  the  common  chord  are  perpendicular 
bisectors  of  each  other. 

Exercise.    Solve  Ex.  1,  §151,  by  Th.  74  Cor. 


TANGENT   CIRCLES 

153.  Theorem  75.    If  two  circles  meet  at  a  point  on  their 
line  of  centers,  the  circles  are  tangent. 


Fig.  213 

Suggestion.  Use  an  indirect  proof.  Suppose  that  they  have  a 
second  point  in  common.     See  Th.  72  and  Th.  65,  §139. 

CoR.  I.  If  the  segment  joining  the  centers  of  two  circles 
is  equal  to  the  sum  of  the  radii,  the  circles  are  tangent 
externally. 

CoR.  IL  If  the  segment  joining  the  centers  of  two  circles 
is  equal  to  the  difference  between  the  radii*  the  circles  are 
tangent  internally. 


120 


PLANE  GEOMETRY 


Theorem  76.     If  two  circles  are  tangent,  the  point  of 
contact  is  on  the  line  of  centers. 

Suggestion.     Use  an  indirect  proof. 

Ex.  1.     In   Fig.  214,   ®A    and  B  are  tangent  at    point   0. 
Show  that  they  have   a  common   tangent, 
OX,  at  point  0.     Construct  OX.  r^ 


Ex.  2.  In  Fig.  214,  XY  and  XZ  are  tan- 
gent to  d)  A  and  B  respectively  from  any 
point  in  the  common  tangent  OX.      Prove  Fig.  214 

that  XY  =  XZ  if  Y  and  Z  are  the  points  of  contact. 

Ex.  3.  In  Fig.  215,  ^  and  B  are  two  equal  circles 
tangent  at  C.  XY  is  the  common  tangent  at  the 
point  C.  Prove  that  any  point  on  XF  may  be  the 
center  of  a  circle  tangent  to  ©A  and  B. 


Y 

Fig.  215 


Ex.  4.  In  Fig.  216,  A  and  B  are  two  equal 
non-intersecting  circles.  XF  is  a  ±  bisector  of  the 
line  of  centers  AB.  Prove  that  any  point  on  XF 
may  be  the  center  of  a  circle  tangent  to®  A  and  B. 


Ex.  5. 


Fig.  216 
Investigate  the  case,  Ex.  4,  if  the  circles  intersect. 


Ex.  6.     Show  how  to  construct  circles  that  will  be  tangent  to 
each  of  two  concentric  circles. 

Ex.  7.     With  three  given  segments  as  radii  construct  three 
circles  each  tangent  to  the  other  two. 


SUPPLEMENTARY  EXERCISES 

EXERCISES  INVOLVING  INSCRIBED  AND 
CIRCUMSCRIBED  POLYGONS 

154.  A  polygon  is  said  to  be  inscribed  in  a  circle  if  its 
vertices  are  on  the  circle  and  its  sides  are  chords  of  the 
circle.  In  this  case  the  circle  is  said  to  be  circumscribed 
about  the  polygon.  A  polygon  is  said  to  be  circumscribed 
about  a  circle  if  its  sides  are  tangent  to  the  circle.  In  this 
case  the  circle  is  said  to  be  inscribed  in  the  polygon. 


CIRCLES   AND   RELATED   LINES  121 

1.  Inscribe  an  equilateral  octagon  in  a  given  circle. 

Suggestion.    It  is  necessary  to  divide  the  pcrigon  at  the  center  of 
the  circle  into  8  equal  parts. 

2.  Inscribe  an  equilateral  hexagon  in  a  given  circle. 

Suggestion.    One-sixth  of   the   perigon  is  60°.     How  are  angles  of 
60°  constructed? 

3.  Inscribe  an  equilateral  polygon  of  12  sides  in  a  given  circle; 
also  one  of  16  sides. 

4.  Prove  that  the  polygons  constructed  in  Exs.  1,  2,  and  3  are 
regular. 

5.  If  a  regular  pentagon  is  inscribed  in  a  circle,  its  diagonals  are 
equal. 

Note.  The  pupil  cannot  construct  a  regular  pentagon  at  present 
without  a  protractor. 

6.  Any  of  the  longer  diagonals  of  a  regular  hexagon  is  a 
diameter  of  the  circumscribed  circle. 

7.  How  many  sets  of  equal  diagonals  has  a  regular  octagon? 
Are  any  of  them  diameters  of  the  circle? 

8.  AB  and  CD  are  two  diameters  of  a  circle  perpendicular  to 
each  other.    Prove  that  tangents  at  their  extremities  form  a  square. 

9.  The  sum  of  two  opposite  sides  of  a  circumscribed  quadri- 
lateral is  equal  to  the  sum  of  the  two  remaining  sides. 

10.  AX  and  BX  are  tangents  to  ©0  And  meet  at  point  X 
without  OO.  AO  and  BO  are  radii  drawn  to  the  points  of 
contact  A  and  B.  Prove  that  Z0-\- ZX  =  2  rt.  A  ;  that  OX  bisects 
ZO  and  ZX  and  is  a  perpendicular  bisector  of  the  chord  AB. 

11.  To  circumscribe  about  a  given  circle  a  triangle 
whose  angles  shall   be   equal  to  three  given  angles. 

Analysis  and  directions: 

I.  To  circumscribe  a  triangle  about  a  circle,  construct 

the  sides  tangent  to  the  circle. 
II.  To  construct  ZA  equal  to  one  of  the  given  angles, 

construct  Z  YOX  equal  to  the  supplement  of  Z^  (Fig.  217). 
Let  the  pupil  complete  the  directions,  construct  the  figure,  and  give 
the  proof. 


122  PLANE   GEOMETRY 

MISCELLANEOUS  EXERCISES 
155.     Note.     Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend'. 

1.  If  two  lines  intersect  at  a  point  within  a  circle  and  make 
equal  angles  with  the  segment  joining  the  point  of  intersection  and 
the  center  of  the  circle,  the  chords  cut  off  are  equal. 

2.  Investigate  the  case,  Ex.  1,  {a)  when  the  point  of  intersection 
is  on  the  circle;  (b)  when  the  point  of  intersection  is  without  the 
circle. 

3.  If  two  equal  chords  intersect  within  a  circle,  they  make  equal 
angles  with  the  segment  joining  the  point  of  intersection  and  the 
center  of  the  circle. 

4.  Investigate  the  case,  Ex.  3,  (a)  when  the  equal  chords  inter- 
sect on  the  circle;  (b)  when  the  equal  chords  are  segments  of  lines 
that  intersect  without  the  circle. 

5.  A  perpendicular  bisector  of  a  chord  bisects  all  chords  parallel 
to  the  given  chord. 

6.  A  line  joining  the  mid-points  of  two  parallel  chords  passes 
through  the  center  of  the  circle. 

7.  If  a  tangent  and^a  chord  are  parallel,  they  cut  off  equal  arcs. 

8.  Two  parallel  chords  in  a  circle  cut  off  equal  arcs. 

Suggestion.    Draw  a  diameter  perpendicular  to  one  of  the  chords  and 
use  the  As. :    Equal  arcs  subtracted  from  equal  arcs  give  equal  arcs. 

9.  A  chord  through  the  point  of  tangency  of  tangent  circles 
subtends  equal  central  angles  in  the  two  circles. 

10.  A  line  through  the  center  of  a  circle  is  cut 
by  two  parallel  tangents.  Prove  that  the  segments 
cut  from  this  line  between  the  tangents  and  the 
circle  are  equal. 

11.  In  Fig.  218,  AB  is  any  chord  and  CD  is  a 
diameter  intersecting  the  chord.  DE  and  CF  are 
±  AB  from  D  and  C  respectively.  Prove  that 
AE  =  BF. 

12.  In  Fig.  219,  ^^  is  any  chord  in  O  0.  CD 
is  a  diameter  drawn  to  the  mid-point  of  arc  AB. 
Prove  /1=Z2. 


CHAPTER  VII 
Circles  and  Related  Angles 

RELATION  BETWEEN   CENTRAL   ANGLES  AND 
THEIR  ARCS 

UNITS  FOR  MEASUREMENT  *OF  ARCS 

156.  There  are  two  ways  by  which  an  arc  is  measured: 
by  its  length  and  by  the  number  of  degrees  it  contains.  In 
this  chapter  we  consider  the  measure  of  an  arc  in  degrees. 
The  number  of  degrees  in  an  arc  is  closely  related  to  the 
number  of  degrees  in  certain  angles. 

The  degree  of  angle  is  Meo  of  a  complete  rotation.  Since 
in  the  same  circle  equal  central  angles  intercept  equal  arcs, 
Heo  of  a  complete  rotation  about  any  point,  as  O,  will  inter- 
cept Keo  of  any  circle  drawn  with  O  as  center.  The  arc 
intercepted  by  the  unit  angle  of  one  degree  at  the  center 
of  a  circle  is  taken  as  a  unit  for  measuring  the  arcs  of  that 
circle.  The  unit  arc  is  therefore  Meo  of  the  circle  and  is 
called  a  degree  of  arc.  Smaller  units  are  obtained  by 
using  smaller  divisions  of  the  dngle.  An  arc  of  one  minute 
(')  corresponds  to  a  central  angle  of  one  minute.  An  arc  of 
one  second  (")  corresponds  to  a  central  angle  of  one  second. 

Ex.  1.  Are  two  angles  of  the  same  number  of  degrees  con- 
gruent? Are  two  arcs  of  the  same  number  of  degrees  always 
congruent?     Illustrate  by  a  figure. 

Ex.  2.  If  a  wheel  makes  250  revolutions  a  minute,  through 
how  many  degrees  does  it  revolve  in  one  second? 

Ex.  3.  If  a  a  A  BCD  is  inscribed  in  a  circle,  how  many 
degrees  in  the  arc  AB? 

123 


124  PLANE  GEOMETRY 

FUNDAMENTAL   RELATION 

167.  It  follows  that  a  central  angle  of  30°  intercepts  an 
arc  of  30°;  a  central  angle  of  42°  7'  15"  intercepts  an  arc  of 
42°  7'  15".  We  shall  accordingly  assume  that  if  the  measure 
of  a  central  angle  is  any  number  whatsoever,  the  measure 
of  the  intercepted  arc  is  expressed  by  the  same  number. 
We  have,  therefore, 

As.  53.  The  measure  of  a  central  angle  and  its  inter- 
cepted arc  are  expressed  by  the  same  number,  or  a  central 
angle  is  measured  by  its  intercepted  arc. 

158.  A  protractor  is  an  instrument  for  measuring  angles. 
It  usually  consists  of  a  semicircle  or  circle  divided  into  unit 
arcs  and  is  used  like  any  other  scale  for  measuring.  Fig.  220 
shows  one  form  of  protractor. 


Ex.  1,    To  measure  an  angle  with  a  protractor, 
Place  the  protractor  with  the  center  C  on  the  vertex  of  the 
angle  and  the  line  of  zeros  CO  along  a  side  of  the  angle.     Read  off 
the  number  of  degrees  on  the  scale  as  indicated  by  the  other  side 
of  the  angle. 

Ex.  2.  Show  how  to  construct  with  the  protractor  an  angle 
that  shall  have  a  given  number  of  degrees. 

Ex.  3.  Construct  with  the  protractor  angles  of  54°,  72°,  125°, 
40°,  18°.    Draw  a  number  of  angles;  measure  each  with  protractor. 

Ex.  4.  With  a  protractor  divide  a  given  circle  into  10  equal 
parts;  into  9  equal  parts;  into  15  equal  parts.  By  this  means 
inscribe  in  a  given  circle  equilateral  polygons  of  10,  9,  and  15 
sides.     Prove  that  these  polygons  are  regular. 


CIRCLES   AND   RELATED  ANGLES  125 

INEQUALITIES   IN   CIRCLES 

169.  The  following  are  the  fundamental  assumptions 
concerning  inequalities  in  circles: 

As.  54.  In  the  same  circle  or  in  congruent  circles,  if 
two  central  angles  are  unequal,  the  minor  arc  subtended  by 
the  greater  angle  is  greater  than  the  minor  arc  subtended 
by  the  lesser  angle. 

As.  55.  In  the  same  circle  or  in  congruent  circles,  if 
two  minor  arcs  are  unequal,  the  angle  subtended  by  the 
greater  arc  is  greater  than  the  angle  subtended  by  the 
lesser  arc. 

fEx.  1.  In  the  same  circle  or  in  congruent  circles,  if  two  chords 
are  unequal,  the  central  angle  subtended  by  the  greater  chord  is 
greater  than  the  central  angle  subtended  by  the  lesser  chord. 

Suggestion.     Draw  radii  to  the  ends  of  the  chords  and  apply  Th.  60. 

fEx.  2.  Investigate  the  converse  of  Ex.  1  and  prove  your 
conclusion. 

fEx.  3.  In  the  same  circle  or  in  congruent  circles,  if  two  minor 
arcs  are  unequal,  the  chord  subtended  by  the  greater  arc  is  greater 
than  the  chord  subtended  by  the  lesser  arc. 

Suggestion.     Apply  As.  55  and  the  preceding  exercise. 

fEx.  4.  Investigate  the  converse  of  Ex.  3.  Prove  your  con- 
clusions. 

fEx.  5.  In  the  same  or  in  congruent  circles  the  greater  of  two 
unequal  chords  is  at  the  less  distance  from  the  center  (Fig.  221). 

A  nalysis: 
I.  To  prove  perpendicular  OY  <    perpendicular 
OX,  prove  OF  <  a  part  of  OX. 

11.    .*.  place  /1 5  with  point  A  on  point  C,  and  so  that 

B  lies  between  D  and  C,  prove  that  OZ  cuts  ^~^ 

DC,  and  that  OF  <  a  part  of  OZ.  .    Fig.  221 

III.  To  place  AB  in  position  required  in  II,  prove  AB  <  DC. 
fEx.  6.     Prove  the  converse  of  Ex.  5  by  an  indirect  proof. 
Ex.  7.     Construct  in  a  given  circle  the  shortest  chord  that 
shall  pass  through  a  given  point. 


126 


PLANE   GEOMETRY 


RELATION   BETWEEN   INSCRIBED   ANGLES 
AND   THEIR  ARCS 

THE  MEASURE   OF  THE   INSCRIBED   ANGLES 

160.  An  angle  is  said  to  be  inscribed  in  a  circle  if  its 

vertex  is  on  the  circle  and  its  sides  are  chords  of  the  circle. 
The  arc  cut  off  between  the  sides  of  an  inscribed  angle 
is  called  its  intercepted  arc. 

Exercise.     Inscribe  in  a  given  circle  angles  of  44°,  72°,  105°, 
etc.     Find  the  measure  of  the  intercepted  arcs.     Use  a  protractor. 

161.  Theorem  77.    An  inscribed  angle  is  measured  by 
one-half  its  intercepted  arc. 


Fig.  222 

Hypothesis:     ZCAB  is  inscribed  in  OO. 
Conclusion:     Z  CAB  is  measured  by  }4  EC. 
Case  A.     When  the  center  of  the  circle  is  on  one  side  of 
the  angle. 

Analysis  and  construction: 
I.  To  prove  that  Z.A  is  measured  by  ^i  BC,  compare 

ZA  with  an  angle  whose  measure  is  known. 
II.    .*.  connect  C  and  0  and  compare  ZA  and  Zx. 
III.  Tocompare  ZAa.nd  Z:^;, compare  Z A -\-  ZC with  Zx^ 
Proof: 

STATEMENTS 

I.   1.  ZA+ZC=  Zx. 

2.  ZA=ZC. 

3.  :.2ZA=Zx. 


ZA 


Zx. 


CIRCLES  AND  RELATED  ANGLES  127 

II.    Zx  is  measured  by  BC. 

,'.  ZA  is  measured  by  J^  BC. 
Let  the  pupil  give  all  reasons. 

Case  B.     When  the  center  of  the  circle  is  within  the  angle. 

Analysis: 
I.  To  prove  that  ZA  is  measured  by  >2  BC,  compare 
Zi4  with  angles  whose  measures  are  known. 


II.    .%  draw  the  diameter  through  point  A  and  compare 
ZA  with  Zx  and  Zy. 

Proof: 

STATEMENTS  REASONS 

1.  Zx  is  measured  by  3^  5X.  L  Why? 

2.  Z;v  is  measured  by  }4  CX.  2.  Why? 

3.  .*.   ZA  is  measured  by  }4  3.  If  equal  numbers  are 
BC.  added  to  equal  num- 
bers, etc. 

Case  C.     When  the  center  of  the  circle  is  without  the 
angle. 

The  analysis  and  the  proof  are  left  to  the  pupil. 

Ex.  1.     If  two  angles  of  an  inscribed  triangle  are  70°  and  50*, 
find  the  number  of  degrees  in  the  arcs  subtended  by  each  side. 

Ex.  2.     The    arcs    subtended    by    the   sides  of   an   inscribed 
triangle  are  in  the  ratio  of  1:2:3.     Find  the  num-         f  e  ^ 
bar  of  degrees  in  each  angle  of  the  triangle.  ^^^r^fT^^^c 

Ex.  3.     In  Fig.  223,  the  semicircle  is  divided  into  / \ 

5  equal  parts  by^the  points  C,  D,  F,  and  G.     E  is  the      ^^^'  ^23 
mid-point  of  FD.     Find  the  nurfiber  of  degrees  in      ^dSign^°° 
each  angle  of  the  figure. 

_Ex.^.  In  Fig.  224,  i^  =  120°  and  j5c=100°. 
AD  =  BC.  Find  the  number  of  degrees  in  each  angle 
of  the  figure. 

Fig.  224 


128  PLANE  GEOMETRY 

COROLLARIES:     TESTS  FOR  EQUAL  ANGLES,  RIGHT  ANGLES 
AND    SUPPLEMENTARY  ANGLES 

162.  Cor.  I.  Inscribed  angles  measured  by  the  same  or 
by  equal  arcs  are  equal,  and,  conversely,  arcs  that  measure 
equal  inscribed  angles  are  equal. 

Ex.  1,  If  the  vertices  of  a  square  A  BCD  lie  on  a  circle  and 
E  is  any  point  in  the  arc  AB,  pt-ove  that  CE  and  DE  trisect  the 
ZAEB. 

An  angle  is  said  to  be  inscribed  in  an  arc  if  its  vertex 
lies  on  the  arc  affti  its  sides  pass  through  the  extremities 
of  the  chord  of  the  arc.  In  this  case  the  arc  is  said  to 
contain  the  angle. 

Cor.  II.  An  angle  inscribed  in  a  semicircle  is  a  right 
angle. 

For  summary  of  tests  for  perpendiculars  and  right 
angles,  see  p.  301. 

Ex.  2.  The  angle  between  the  segments  joining  a  point  with- 
in a  circle  to  the  ends  of  a  diameter  is  an  obtuse  angle. 

Ex.  3.  The  angle  between  the  segments  joining  a  point  with- 
out a  circle  to  the  ends  of  a  diameter  is  an  acute  angle. 

Ex.  4.  If  the  radius  of  one  circle  is  the  diameter  of  a  second, 
any  chord  of  the  larger  drawn  from  the  point  of  contact  is  bisected 
by  the  smaller  circle.  ^ 

Construct  a  right  triangle,  given 

Ex.  5.     The  hypotenuse  and  one  leg. 

Ex.  6.     The  hypotenuse  and  an  acute  angle. 

CoR.  III.  Inscribed  angles  are  supplementary  if  the 
sum  of  their  intercepted  arcs  is  360°. 

Ex.  7.  The  opposite  angles  of  an  inscribed  quadrilateral  are 
supplementary. 

Ex.  8.  If  a  triangle  is  inscribed  in  a  circle,  the  sum  of  the 
angles  inscribed  in  the  arcs  subtended  by  the  sides  is  4  right  angles. 


CIRCLES  AxND   RELATED  ANGLES 


129 


EXERCISES  INVOLVING  CENTRAL  AND  INSCRIBED  ANGLES 

163.  The  foregoing  theorem  and  its  corollaries  give  us 
L  A  new  method  for  proving  angles  equal. 
XL  A  new  method  for  proving  arcs  equal. 

III.  A  new  method  for  proving  angles  supplementary. 

IV.  A  new  method  for  determining  and  constructing  right 

angles. 

1.  Find  the  arc  which  measures   ZACD  in 
Fig.  225. 

2.  If  from  point  P  on  a  circle  PQ  and  a  b\ 
diameter   PR   are   drawn,  a  radius  parallel  to 
PQ  bisects  QR.  Fig.  225 

3.  A  circle  is  circumscribed  about  A  ABC  and  P  is  the  mid- 
point of  AB.    Prove  that  ZABP  is  K  /C. 

4.  Construct  a  number  of  angles  inscribed  in  the  same  arc  and 
the  bisector  of  each  angle.     What  theorem  can  be 
inferred  from  the  drawing?     Prove  it. 

5.  Show  that  the  center  of  a  given  circle  may  be 
found  with  a  carpenter's  steel  square  as  shown  in 
Fig.  226.     Give  reasons  (Th.  50). 

6.  ABC  is  a  triangle  inscribed  in  a  circle  whose 
center  is  0.  OD  is  perpendicular  to  AC.  Prove 
that   ZCOD=ZB. 

7.  Show  that  the  center  of  a  given  arc  may  be 
found  by  using  two  carpenter's  steel  squares  as  shown 
in  Fig.  227.    Give  reasons. 

8.  In    Fig.    228,  ©0  and   0'   are  two  equal     ^" 
circles  intersecting  at  A  and  B.     C  and  D  are  any 
points  on  CD  0  and  0'  respectively.    Prove  Z  BCA 
=  ZADB.  Fig.  228 

9.  A  central  angle  is  twice  an  inscribed  angle  that  intercepts 
the  same  arc. 

10.  Inscribe  in  a  given  circle  a  triangle  whose  angles  are  equal 
to  the  angles  of  a  given  triangle.     Suggestion.    Use  Ex.  9. 


130 


PLANE  GEOMETRY 


ANGLES   FORiVLED   BY  INTERSECTING   CHORDS 

164.  Theorem  78.  An  angle  formed  by  two  chords 
intersecting  within  a  circle  is  measured  by  one-half  the 
sum  of  the  intercepted  arcs. 


Fig.  229 

Hypothesis:    The  chords  AC  and  BD  intersect  within 
OO. 

Conclusion:     Zl  is  measured  by  '^  {AD-{-BC). 

Analysis  and  construction: 

I.  To  prove  that  Z  1  is  measured  by  }4  {AD -{-EC),  com- 
pare Z  1  with  angles  whose  measures  are  known. 

il.    .*.  connect  A  and  B  and  prove  Zl=  Z.A-\-  ZB. 

Let  the  pupil  give  the  proof. 

_Ex.  l._  If,  in  Fig.  230,  ^  =  55°,  ^'S=140°;  and 
BC=%  AD,  find  the  number  of  degrees  in  each  angle 
of  the  figurCo 

Ex.   2.     Discuss   the  special  case   of  Th.  78  in 
which  the  given  chords  intersect  at   the  center  of 

the   circle. 

• 

Ex.  3.     In  Fig.  231,  P  is  the  mid-point  of  CD  A 
PA  and  PB  are  any  two  chords  from  P.     Prove  that 
Zl=Zl'and   Z2=Z2'.  p,^  231 

Ex.  4.  Find  the  sum  of  each  pair  of  opposite  arcs  into  which 
two  perpendicular  chords  divide  a  circle. 

Ex.  5.  If,  in  Fig.  230,  AC  had  been  a  diameter,  Z  1  had  been 
72°,  and  DC  had  been  h  AB,  what  would  have  been  the  number  of 
degrees  in  each  of  the  other  angles  of  the  figure? 


CIRCLES  AND  RELATED  ANGLES 


131 


ANGLES   FORMED   BY   SECANTS   INTERSECTING 
WITHOUT   THE   CIRCLE 

165.  A  line  of  indefinite  length. which  cuts  a  circle  in  two 
places  is  called  a  secant. 

Theorem  79.  An  angle  formed  by  two  secants  inter- 
secting without  a  circle  is  measured  by  one-half  the  differ- 
ence of  the  intercepted  arcs. 


Fig.  232 

Hypothesis:     The  secants  BA  and  BC  intersect  without 

oo.  ■ 

Conclusion:     Z.B  is  measured  by  ^4  {AC—XY). 
Analysis  and  construction: 
I.  To  prove   /LB  measured  by  >^  {AC-XY),  compare 
'   AB  with  angles  whose  measures  are  known. 

II.    .*.  connect  C  and  X  and  prove  AB=  Z.AXC—  AC. 

III.    .-.  prove  AAXC  =  ZB  +  ZC. 

Let  the  pupil  give  the  proof. 

Ex.  1.  In  Fig.  233,  secants  a  and  b  meet 
without  OO.  If  secant  b  moves  until  it 
comes  into  the  positions  b'  and  b'\  how  are 
41,  2,  and  3  measured?  Give  the  theorem 
that  applies  in  each  case. 

Ex .  2.  If,  in  Fig.  234,  Z  C  =  25°  and  XV  = 
H  A  B,  find  the  number  of  degrees  in  Z  XA  Y 
and  in  ZBYA. 

Ex.  3.     If,  in  Fig.  234,  ZC  =  24°, XY^VsAB,^ 
and  AV  is  3.  diameter,  find  the  number  of  degrees 
in  each  angle  of  the  figure. 


Fig.  233 


Fig.  234 


132  PLANE   GEOMETRY 

ARCS   FORMED   BY   PARALLEL   CHORDS:     TEST 
FOR   EQUAL   ARCS 

166.  Theorem  80.     Parallel  chords  intercept  equal  arc: 
on  a  circle. 


Fig.  235 

Hypothesis:     OO  is  any  circle  and  AB  ||  DC. 
Conclusion:    AD  =  BC. 
Analysis  and  construction: 

I.  To  prove  AD  =  BC,  prove  that  they  are  intercepted  by 
equal  inscribed  angles. 

II.    .*.  join  AC 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 

_Ex.  1.     If,  in  Fig.  235,  AD  =  32°  and  minor  DC  is  H  of  minor 
AB,  find  the  number  of  degrees  in  each  arc. 

Ex.  2.     If  on  a  circle  AC  =  BD,  prove  that  AD  is  either  par- 
allel with  or  equal  to  BC. 

Ex.  3.     If  the  vertices  of  a  trapezoid  lie  on  a  circle,  its  diago- 
nals are  equal  and  its  base  angles  are  equal. 


Ex.  4.     Prove  Th.   79   by   comparing    ZB 
with  one  inscribed  angle. 

Suggestion.     From  D  construct  a  line  parallel 
toAB  and  compare  ZB  with  ZFDC  (Fig.  236). 


Fig.  236 


Ex.  5.     Prove  Th.  78  by  comparing   Zl   with  one  inscribed 
angle. 

Suggestion.     From  A  construct  a  line  ||  DB. 


CIRCLES  AXD  RELATED  ANGLES 


133 


CONSTRUCTION    OF   PERPENDICULARS 

167.  Probleim  9.     To    construct    a    perpendicular    to    a 
line  from   a  point   in   the   line. 


/!->. ^-'' 

Fig.  237 

Analysis  and  constniction: 
I.  In  order  to  construct  a  perpendicular  to  line  I  at  A, 
construct  a  right  angle  at  A . 
II.  To  construct  a  right  angle  at  A,  inscribe  in  a  semi- 
circle an  angle  with  its  vertex  at  A. 
III.  To  construct  a  semicircle  through  A,  take  any  point  O 

as  center  and  segment  OA  as  radius 

Let  the  pupil  complete  the  directions  and  give  the  proof. 

168.  Problem  10.     To   construct   a   perpendicular   to   a 
line  from  a  point  not  in  the  line. 


^^ 

r 

/ 

/ 

1 

1 

^^ 

\     / 

^^ 

L )^ — 

I 

Fig.  238 
Analysis  and  construction: 
I.  In  order  to  construct  a  perpendicular  to  line  /  through 
point  A,  construct  a  right  angle  with  its  vertex 
on  /  and  with  one  side  passing  through  point  A. 
*.  construct  a  circle  through  A  cutting  Hne  /. 
*.  use  any  oblique  segment  through  A  to  line  /  as 
a  diameter. 


II 
III 


Let  'the  pupil  complete  the  directions  and  give  the  proof. 
10 


134  PLANE  GEOMETRY 

CONSTRUCTION   OF   TANGENTS 

169.  Problem  11.     To  construct  a  tangent  to  a  circle 
from  a  given  point  without  the  circle. 


Fig.  239 

Given  QO  and  point  A  without  the  circle. 
To  construct  sl  tangent  to  OO  from  point  A. 
Analysis  and  construction: 
I.  In  order  to  construct  a  tangent  to  OO  from  point  Ay 
construct  a  right  angle  whose  vertex  is  on    OO 
and  whose  sides  pass  through  points  O  and  A. 
II.    /.  join  OA  and  construct  a  circle  on  OA  as  diameter, 

cutting  OO  at  X  and  Y.  Join  AX  and  AY. 
Let  the  pupil  give  the  proof. 
Ex.  1.  Let  P  represent  any  point  within 
OO  (Fig.  240).  Suppose  P  to  move  away 
from  the  center  along  the  ray  OX.  How 
many  tangents  can  be  drawn  to  OO  through 
point  P  when  P  is  within  the  circle?  when 
P  is  on  the   circle?  when  P  is  outside  the  pj^,^  240 

circle?  How  are  these  tangents  constructed 
in  each  case?  Locate  various  positions  of  P  outside  the  circle 
and  sketch  in  the  tangents  from  P  at  each  position  of  P.  As  P 
moves  away  from  the  circle  how  does  the  angle  between  the 
tangents  change?  How  do  the  points  of  contact  move?  What 
are  the  limiting  positions  of  these  points?  How  does  the  length 
of  the  segment  between  P  and  the  point  of  contact  change?  What 
are  the  limiting  lengths  of  this  segment? 

Ex.  2.     Circumscribe  an  isosceles  triangle  about  a  circle,  given 
the  base  of  the  triangle.     Is  the  problem  always  possible? 


CIRCLES  AND  RELATED  ANGLES 


135 


RELATION    BETWEEN    ANGLES    FORMED   BY 
TANGENTS  AND   CHORDS,  AND  THEIR  ARCS 

MEASUREMENT   OF  THE   ANGLE 

170.  Theorem  8L    An  angle  formed  by  a  tangent  and 
a  chord  is  measured  by  one-half  its  intercepted  arc. 


Fig.  241 

Hypothesis:    AB  is  tangent  to  QO  oX  A  and  ^C  is  drawn 
from  A ,  the  point  of  contact  of  the  tangent. 

Conclusion:     Z  1  is  measiired  by  ^  AC. 

Analysis  and  construction: 

I.  To  prove  that  Zl  is  measured  by  >^  AC,  compare 
Z  1  with  angles  whose  measures  are  known. 

II.    .*.  draw  diameter  AD  and  compare    Zl  wdth    Z3 
and  Z2. 

III.  To  find  the  measure  of    Z3,  prove    Z3  =  90°  and 

ACD  =  \m\ 
The  proof  is  left  to  the  pupil. 
Ex.  1.     In    Fig.    241,  prove    that     ABAC    Is    measured  by 
K  ADC. 

Ex.2.     Prove  Th.  81  by  comparing    A  CAB 
with  one  inscribed  angle. 
Suggestion.     Draw  CD. 

Ex.3.     Prove  Th.  81   by  comparing   A  CAB 
with  a  central  angle  (Fig.  242).  Fig.  242 


136 


PLANE  GEOMETRY 


ARCS   FORMED   BY   PARALLEL   CHORD   AND   TANGENT 

171.  Theorem  82.    If  a  chord  and  a  tangent  are  par- 
allel, they  cut  off  equal  arcs. 

The  analysis  and  the  proof  are  left  to  the  pupil. 

Exercise.     The  chord  of  an  arc  is  parallel  to  the  tangent  drawn 
to  the  mid-point  of  the  arc. 


ANGLES   FORMED   BY   SECANTS  AND   TANGENTS 

172.  Theorem  83.  An  angle  formed  by  a  secant  and  a 
tangent  is  measured  by  one-half  the  difference  of  the  inter- 
cepted arcs. 

Theorem  84.  An  angle  formed  by  two  tangents  is 
measured  by  one-half  the  difference  of  the  intercepted  arcs. 

Let  the  pupil  review  Ths.  78  and  79. 

Ex.  1.  In  Fig.  243,  if  AXC  is  220°,  find 
the  number  of  degrees  in  ZB. 

Ex.  2.  In  Fig.  243,  if  Z  5  is  68°,  find  the 
number  of  degrees  in  AXC  and  A  YC. 

Ex.  3.  Two  tangents  are  perpendicular 
to  each  other.  How  many  degrees  in  the  arcs 
formed  by  the  points  of  tangency? 

Ex.  4.  If  line  BC  (Fig.  243)  moves  so  that  it  remains  parallel 
to  its  original  position,  how  will  ZB  he  measured  when  BC  passes 
through  point  A? 

Ex.  5.  In  Fig.  244  let  AB  and  CD 
represent  any  two  chords  intersecting 
within  O  0.  How  is  Z 1  measured? 
How  will  the  measure  of  Z  1  change  if 
CD  revolves  about  point   C  in  either  Fig.  244 

direction?  Various  positions  of  CD  are  shown  by  the  dotted 
lines.  How  would  Zl  be  measured  in  each  case?  Discuss  the 
vase  in  which  CD  becomes  parallel  to  AB. 


CIRCLES  AND  RELATED  ANGLES  137 

SUMMARY  AND   SUPPLEMENTARY  EXERCISES 
173.  SUMMARY   OF   IMPORTANT   POINTS   IN    CHAPTER   VII 
I.  Tests  for  measurement  of  angles. 

a.  A  central  angle  is  m'easured  by,  etc.  (§157). 

b.  An  inscribed  angle  is  measured  by,  etc.  (§161). 

c.  An  angle  formed  by  a  tangent  and  a  chord,  etc. 

(§170). 

d.  An   angle   formed   by   two   chords   intersecting 

within,  etc.  (§164). 

e.  An  angle  formed   by   two  secants  intersecting 

without,  etc.  (§165). 
/.   An  angle  formed  by  a  secant  and  a  tangent 

(§172). 
g.  An  angle  formed  by  two  tangents,  etc.  (§172). 

II.  General  test. 

Angles  and  arcs  are  both  measured  in  degrees.  It  is  evi- 
dent, therefore,  that  in  the  same  circle  or  in  congruent 
circles : 

Angles  are  equal  if  they  are  measured  by  equal  arcs. 

Arcs  are  equal  if  they  measure  equal  angles. 
For  summary  of  test  of  equal  arcs,  see  p.  303. 

EXERCISES  CONCERNING  TANGENTS  TO  CIRCLES 

174.     1.  Prove  that  a  tangent  can  be  drawn  to  a  circle  from 

a  given  point  without  the  circle  by  the  following  method   (Fig. 

24 r>) :  Let  yl  be  the  given  point.    Join  0/1.     With        ,-- — ^-* 

0  as  center  and  OA  as  radius  draw  circle  ABC.    /  ^ — ^  '> 

Let  OA  cut  the  given  circle  at  D.    At  D  draw  ;'     [     o^^l.^P^^ 

BC  tangent  to  the  given  circle  and  ciitting  the   \    V      ^V>4/' 

outer  circle  at  B  and  C.     Draw  OB  and  OC  cut-     '\^         ^^'l^' 

ting  the  given  circle  at  E  and  F.     AE  and  AF  "*""!' 

i.  X     XI.       •  -1  Fig.  245 

are  tangent  to  the  given  circle. 

Suggestion.  Prove  Z1=Z2  =  1  rt.  Z,  by  comparing  AOEA  with 
AODB. 

Note.     The  construction  given  in  Ex.  1  is  similar  to  Euclid's. 


138  PLANE  GEOMETRY 

t2.  From  the  following  figures   (Fig.  246)   give  the  analysis, 
directions,    and    proof    for    constructing  the  common  tangents 


-V— , 


Fig.  246 

to  two  given  circles.     Note  the  application  of  this  to  belts  over 

pulleys.  •  . 

3.  A  line  that  is  tangent  to  each  of  two  equal  circles  is  parallel 
to  the  segment  joining  their  centers  or  it  bisects  this  segment. 

4.  Two  common  interior  tangents  to  two  circles  are  equal. 

5.  Is  the  same  true  of  the  exterior  tangents? 

6.  How  many  common  tangents  can  be  drawn  to  two  circles 
if  they  are  in  each  of  the  possible  positions  shown  in  Fig.  209? 
Show  how  to  construct  the  tangents  in  each  case. 

7.  An  angle  between  two  tangents  to  a  circle  is  double  the 
angle  between  the  chord  joining  the  point  of  contact  and  the 
radius  drawn  to  one  point  of  contact.  ^ 

8.  Circumscribe  an  isosceles  triangle  about  a  circle,  given  the 
altitude  of  the  triangle.     Is  the  problem  always  possible? 

9.  Circumscribe  about  a  given  circle  a  right  triangle,  given 
one  leg.     Is  the  problem  always  possible? 

MISCELLANEOUS   EXERCISES 

175.  Note.     Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend. 

1.  Make  review  diagrams  for  Ths.  82  and  83. 

2.  In  Fig.  247  the  sides  of  AXFZ  are  tan- 
gent to  the  circle  at  the  vertices  of  the  inscribed 
triangle  ABC.     If  Zl=43°  and  Z3  =  62°,  find^  ^ 
the  number  of  degrees  in  each  angle  of  the  figure.          Fig.  247 


CIRCLES  AND  RELATED  ANGLES 


139 


3.  If,  in  Fig.  248,  ZCAB  =  5S°,  find  the  number  of  degrees  in 
ZD,  ZE,  and  ZC^F  if  BF  is  given  tangent  to  the  circle. 

4.  A  chord  forms  equal  angles  with  the  tan- 
gents at  its  extremities.  Discuss  the  special  case 
in  which  the  chord  is  a  diameter  of  the  circle. 

5.  Prove  Ths.  83  and  84  bv  a  method  similar  ^ 
to  that  suggested  for  Th.  79  in  Ex.  4,  §  166. 


6.  In  Fig.  249,  AC  is  tangent  to  OO  at  ^. 
Prove  that  Z  CAB  is  equal  to  K ^ 0. 


7.  A  tangent  at  the  vertex  of  an  inscribed  angle  forms  equal 
angles  with  the  sides  of  the  given  angle  if  these  sides  are  equal. 

8.  If  an  isosceles  triangle  is  inscribed  in  a  circle,  the  tan- 
gent at  the  vertex  makes  equal  angles  with  the  legs  and  is 
parallel  to  the  base. 

9.  In  Fig.  250,  AB  equals  CD  and  the  chords 
are  produced  to  intersect  at  P.  Prove  that  the 
segment  PA  equals  the  segment  PD. 

10.  If  two  chords  intersect  in  a  circle  and  a  seg- 
ment of  one  is  equal  to  a  segment  of  the  other, 
the  chords  are  equal.  Fig.  250 

11.  State  and  investigate  the  converse  of  Ex.  10.  Prove  your 
conclusions. 

12.  P  and  Q  are  the  points  of  intersection  of  two  arbitrary 
circles.  PA  and  PB  are  the  diameters  through  P.  Prove  that 
AB  passes  through  Q. 

13.  A    circle   constructed   on   side  ^B  of  A  ABC 
as  diameter  passes  through  the  feet  of  the  perpen-  ^ 
diculars  from  A  and  B  to  the  sides  BC  and  AC  re- 
spectively (Fig.  251).     See  Th.  50.  Fig.  25; 


140 


PLANE  GEOMETRY 


14.  Circles  constructed  on  two  sides  of  a  triangle  as  diameters 
will  intersect  on  the  third  side. 

15.  A  circle  constructed  on  one  leg  of  an  isosceles  triangle  as  a 
diameter  passes  through  the  mid-point  of  the  base. 

16.  If  semicircles  are  constructed  on  the  sides 

of  an  equilateral  triangle  as  diameters,  they  will 

intersect  at  the  mid-points  of  the  opposite  sides     at-^—^  - 

(Fig.  252).  FiG.'252 

Church  window 
detail 

17.  The  ray  which  bisects  the  angle  formed  by  a  tangent  and 
a  chord  bisects  also  the  intercepted  arc. 

18.  If  a  tangent  is  drawn  to  a  circle  at  the  extremity  of  a 
chord,  the  mid-point  of  the  intercepted  arc  is  equally  distant  from 
the  chord  and  the  tangent. 

19.  If  the  extremities  of  any  two  diameters  in  a  circle  are 
joined  in  order,  the  figure  formed  is  a  rectangle. 

20.  If,  in  Fig._^53,  (S)_0  and  O'  are  tan- 
gent at  X  and  AB  and  CD  pass  through  the 
point  of  contact,  prove  that  ^C  is  parallel 
to  BD. 

Fig.  253 

21.  Prove  Ex.  20  if  the  circles  are  tangent  internally. 

22.  If,  in  Fig^  254,  (D  0  and  0'  are  tan- 
gent at  X  and  A  B  passes  through  point  X,  prove 
that  tangents  at  A  and  B  are  parallel. 

Fig.  254 

23.  Prove  Ex.  22  if  the  circles  are  tangent  internally. 

24.  In  Fig-  255,  ©  O  and  0'  are  tangent 
at  X  and  AB  passes  through  X.  Prove  that 
diameters  from  A  and  B  are  parallel. 

Fig.  255 

25.  Prove  Ex.  24  if  the  circles  are  tangent  internally. 


CIRCLES  AND   RELATED  ANGLES 


141 


Fig.  256 


Fig.  257 


26.  In  Fig.  256,  CD  0  and  X  are  tangent  at  C.     AB  is  one 
of  the  common  exterior  tangents  with  A  and  B 
the    points    of    contact.     Prove    that    ZACB 
is  a  right  angle. 

27.  If    two    equal    circles    intersect   and    a 
segment  is  drawn  through  either  point  of  inter- 
section terminating   in  the  circumferences,  the 
segments  joining  the  extremities  with  the   other   point  of  inter- 
section of  the  circles  are  equal. 

28.  In  Fig.  257,  AC  and  BD  are  tangent  to 
OO  at  opposite  ends  of  the  diameter  AB.  CD 
is  an  arbitrary  tangent  intersecting  AC  and  BD 
at  C  and  D  respectively.  Prove  that  /.COD  is  a 
right  angle. 

Suggestion.  If  X  is  the  point  of  contact  of  tangent 
CD,  draw  OX,  OC,  and  OD  and  prove  Z2-f  Z3  = 
}/i  the  straight  angle.      .'.  prove  Z  1  =  Z  2  and  Z  3  =  Z  4. 

29.  Given   a  circle  (Fig.  258)  divided  into  eight  equal   parts 
and  the  points  joined  as  indicated, 
prove 

a.  AK  =  KB  =  BYr 

b.  /.A  =  ZB=ZC. 

c.  AK  =  KP. 

d.  OPQRSTU Visa regulsLT    «f 
octagon. 

e.  WXYZ  is  a  square. 

30.  In  Fig.  258  find  the  num- 
ber of  degrees  in  ZA,  ZAKB, 
ZHVB,  ZBVC. 

31.  Construct  a  figure  similar  to  Fig.  258  by  dividing  the 
circle  into  sixteen  equal  parts  and  joining  every  seventh  point  of 
division,  or  by  dividing  the  circle  into  twelve  equal  parts  and 
joining  every  fifth  point,  and  find  the  number  of  degrees  in  the 
angles  formed. 

Note.     Stars  similar  to  the  above,  though  often  more  complicated, 
are  common  in  cut-glass  designs. 


142 


PLANE  GEOMETRY 


32.  Two  circles  intersect  at  the  points  A  and  B.  Through  A 
a  variable  secant  is  drawn  cutting  the  circles  in  C  and  D.  Prove 
that  the  angle  CBD  is  constant  for  all  positions  of  the  secant. — 
College  Entrance  Examination  Board,  Plane  Geometry  Examina- 
tion, 1908. 

33.  In  Fig.  259  the  circle  is 
divided  into  eight  equal  parts. 
Find  the  center  of  AOC  so  that  it 
shall  pass  through  the  three  points 
A,  0,  and  C.  Prove  that  AOC  is  a 
semicircle. 

34.  In  Fig.  259  prove  that  L  is 

the  mid-point    of   OLA    and   OLB. 

Prove  that  OL  and  OM  are  equal; 

also  AL,  LB,  and  BM.  ^     ^,,, 

'         '  Fig.  259 

35.  Do  you  know  any  practical  uses  of  rosettes  similar  to 
the  above? 

36.  A  quadrilateral  A  BCD  is  inscribed  in  a  circle.  At  the 
points  of  division  tangents  are  drawn  forming  a  circumscribed 
quadrilateral.  A^  =  o5°,  BC=UO°,  CS  =  35°.  Find  the  number 
of  degrees  in  each  angle  of  the  inscribed  quadrilateral,  in  each 
angle  of  the  circumscribed  quadrilateral,  and  in  the  angles  formed 
by  the  diagonals  of  each  quadrilateral. 

37.  Fig.  260  shows  a  simple  turnout  used  on 
street  railroad  tracks.  The  rails  AE  and  CF  are 
arcs  of  circles  tangent  to  rails  AB  and  CD  at  A 
and  C  respectively.  Rail  AE  crosses  rail  CD  at  H. 
Prove  that  Zl  made  by  rail  CD  and.  the  tangent 
to  ^E  at  H  is  equal  to  Z2.  O  is  the  center  for 
AE  and  CF. 

^  38.  Fig.  261  shows  two  opposite  turnouts 
from  the  end  of  a  straight  track.  The  curved 
rails  are  tangent  to  the  straight  ones  at  C  and  D 
respectively.  Prove  that  the  angle  between  the 
tangents  at  F  is  equal  to  ZX  +  ZY  ii  X  and  Y  |' 

are  the  centers  for  the  arcs.  Fig.  261 


Fig.  260 


CHAPTER  VIII 

Loci 

GENERAL   CONSIDERATIONS 

DEFINITIONS 

176.  The  exercises  below  illustrate  the  following  defini- 
tions : 

A  point  which  moves  so  as  to  fulfill  some  given  require- 
ment is  called  a  variable  point. 

The  path  of  a  point  which  moves  so  as  to  fulfill  some 
given  requirement  is  called  a  locus. 

A  line  or  group  of  lines  is  called  a  locus  if  they  contain 
all  points  which  fulfill  some  given  requirement  and  contain 
no  other  points. 

Ex.  1.  Find  a  point  which  is  2  in.  from  a  fixed  point  O.  Is 
there  more  than  one  such  point?  If  you  consider  such  a  point 
as  so  moving  that  it  shall  always,  remain  2  in.  from  0,  what  will 
be  its  path?     Is  this  path  located  definitely?     How? 

Ex.  2.  If  A  and  B  are  two  fixed  points,  find  a  point  which  is 
as  far  from  A  as  from  B.  How  many  such  points  can  you  find? 
What  is  the  path  of  a  point  moving  so  that  it  is  always  equally 
distant  from  two  fixed  points?     Is  this  path  a  fixed  path? 

TO   FIND   THE  LOCUS 

177.  In  some  cases  the  line  or  set  of  lines  that  make  up 
the  locus  can  be  found  directly  from  a  knowledge  of  the 
geometry  involved.  In  other  cases  the  locus  may  be  found 
by  locating  several  positions  of  the  moving  point.  If 
positions  enough  are  located,  the  locus  may  often  be  inferred 
from  them.  This  is  a  method  of  discovery  that  is  common 
to  all  scientific  inquiry.     The  correctness  of  the  inference 

143 


144  ,  PLANE   GEOMETRY 

is  then  determined  by  a  careful  demonstration,  the  nature 
of  which  will  be  discussed  in  the  next  section. 

Note.  One  or  more  of  the  following  exercises  may  be  introduced 
here  if  the  teacher  desires.  In  many  cases  the  locus  cannot  be 
found  unless  a  very  large  number  of  points  are  located.  Some  of 
these  exercises  lead  to  other  kinds  of  lines  than  the  straight  line  and 
the  circle,  lines  that  are  studied  in  more  advanced  courses  in  geometry. 
In  these  latter  exercises  the  locating  of  the  points  must  be  by  experi- 
ment, as  the  pupil  knows  no  construction  by  which  he  can  find  them. 
Such  exercises  are  valuable  in  that  they  force  home  to  the  pupil  the 
possibility  and  the  occasional  necessity  of  finding  loci  by  experiment 
and  in  that  they  set  him  to  thinking  on  a  subject  entirely  new  to  him. 
No  proofs  should  be  required. 

EXERCISES   IN   FINDING   LOCI 

Find  the  following  loci  by  experiment. 

1.  The  ray  a  starts  from  point  P  on  O  X  as 
origin.  Find  the  locus  of  the  mid-points  of  the 
chords  cut  from  this  ray  by  the  circle  as  the  ray 
moves  about  point  P  (Fig.  262).  Fig.  262 

2.  Find  the  locus  called  for  in  Ex.  1  if  the  origin  of  a  is  without 
OX. 

3.  A  ladder  stands  upright  against  a  wall.  Find  the  locus  of  a 
point  on  the  middle  round  if  the  foot  of  the  ladder  is  pulled  out 
until  the  ladder  is  flat  on  the  ground. 

4.  OX  and  OY  are  two  lines  at  right  angles  to  each  other. 
Find  the  locus  of  a  point  which  is  twice  as  far  from  OX  as  from  OY. 

5.  In  Fig.  263,  ABCD  is  a  parallelogram 
and  /  any  line.  Imagine  line  /  to  move  so  as 
always  to  remain  parallel  to  its  original  position. 
Find  the  locus  of  the  mid-points  of  the  segments 
cut  from  /  by  the  sides  of  the  parallelogram.  Fig.  263 

6.  Find  the  locus  of  a  point  }4  the  way  up  the  ladder  mentioned 
in  Ex.  3. 

7.  Find  the  locus  of  centers  of  circles  tangent  to  each  of  two 
given  circles  when  (1)  the  one  circle  is  within  the  other,  (2)  the 
circles  intersect,  (3)  one  circle  is  without  the  other,  (4)  the 
circles  are  tangent  internally,  (5)  the  circles  are  tangent  externally. 


LOCI  145 

8.  Find  the  locus  of  centers  of  circles  which  pass  through  a 
given  point  and  are  tangent  to  a  given  line  when  (1)  the  point  is 
on  the  given  line,  (2)  the  point  is  not  on  the  given  Hne. 

9.  Find  the  locus  of  centers  of  circles  which  are  tangent  to 
a  given  circle  and  to  a  given  line  when  (1)  the  line  intersects  the 
circle,  (2)  the  line  is  tangent  to  the  circle,  (3)  the  line  is  wholly 
without  the  circle. 

What  is  the  locus  of: 

10.  A  point  on  the  knob  of  a  swinging  door? 

11.  A  point  on  the  side  of  a  spinning  top? 

12.  The  hub  of  a  wheel  of  a  moving  bicycle? 

13.  A  point  on  the  tire  of  a  moving  wheel?- 

14.  A  point  on  the  rim  of  a  plate  moving  about  another  plate? 
Note.     No  drawings  are  required  for  Exs.  10-12. 

COMPLETE   PROOFS   FOR   LOCI 

178.  After  the  locus  has  been  found  by  some  method,  it 
is  necessary  to  give  a  complete  formal  proof  that  the  line, 
or  set  of  lines,  found  really  constitutes  the  locus.  It  is 
necessary  to  show,  therefore,  that  every  point  on  the  line 
or  set  of  lines  is  a  possible  position  of  the  variable  point 
and  that  every  possible  position  of  the  variable  point  is 
on  this  line  or  set  of  lines.  If  these  two  facts  can  be  proved, 
it  is  evident  that  the  line  or  set  of  lines  constitutes  the  whole 
locus  and  nothing  but  the  locus. 

To  put  it  more  formally: 

In  order  to  prove  that  a  line  or  set  of  lines  is  the  locus 

of  a  point  which  moves  so  as  to  fulfill  certain  requirements, 
prove 

I.  Every  point  in  the  line  or  set  of  lines  fulfills  the 
requirements,  and 

II.  Every  point  which  fulfills  the  requirements  is  on  the 
line  or  set  of  lines. 

Note.     It  is  immaterial  in  which  order  I  and  II  are  proved. 


146  PLANE  GEOMETRY 

LOGI  OF  POINTS 

THE  BISECTOR   OF   AN   ANGLE 

179.  Theorem  85.     The  bisector  of  an  angle  is  the  locus 
of  points  equally  distant  from  the  sides  of  the  angle. 


Fig.  264 


Hypothesis:    AX  bisects  Z  CAB. 

Conclusion:     ^X  is  the  locus  of  points  equally  distant 
from  AB  and  AC;  that  is, 

^     I.  Every  point  in  AX  is  equally  distant  from  AB  and  AC. 

IL  Every  point  that  is  equally  distant  from  AB  and  AC 
lies  in  AX. 

The  analysis  and  the  proof  for  I  are  left  to  the  pupil. 
Analysis  and  constriction  for  II: 

1.  Let  Q  be  a  point  equally  distant  from  AB  and  AC. 

2.  To  prove  that  Q  lies  in  AX,  join  Q  and  A  and  prove 

that  QA  and  AX  coincide. 

3.  To  prove  that  QA  and  AX  coincide,  show  that  both 

bisect  /.A. 
Let  the  pupil  complete  the  analysis  and  give  the  proof. 

Ex.  1.     Can  you  determine  a  point  equally  distant  from  the 
sides  of  an  angle  without  constructing  the  bisector  of  the  angle? 

Ex.  2.     What  is  the  locus  of  a  point  equally  distant  from  two 
intersecting  lines? 


LOCI  147 

THE  PERPENDICULAR  BISECTOR   OF  A  SEGMENT 

180.  Theorem  8G.  The  perpendicular  bisector  of  a  seg- 
ment is  the  locus  of  a  point  equally  distant  from  the  ends 
of  the  segment. 


Hypothesis:  OP  is  the  perpendicular  bisector  of  the 
segment  AB. 

Conclusion:  OP  is  the  locus  of  a  point  equally  distant 
from  A  and  B ;  that  is, 

I.  Every  point  in  OP  is  equally  distant  from  A  and  B, 

II.  Every  point  that  is  equally  distant  from  A  and  B  lies 
in  OP. 

The  analysis  and  the  proof  for  I  are  left  to  the  pupil. 

Analysis  and  construction  Jor  II: 

1.  Let  0  be  a  point  such  that  QA=QB. 

2.  To  prove  that  Q  lies  in  OP,  join  Q  and  0,  the  mid-point 

of  AB,  and  prove  that  QO  and  OP  coincide. 

3.  To  prove  that  QO  and  OP  coincide,   show  that  both 

QO  and  OP  are  perpendicular  bisectors  oi  AB. 
Let  the  pupil  complete  the  analysis  «and  give  the  proof. 

181.  Cor.  If  two  points  are  each  equally  distant  from  the 
extremities  of  a  segment,  the  line  passing  through  these 
points  is  the  perpendicular  bisector  of  the  segment. 

Suggestion.  Show  that  the  two  given  points  must  both  lie  on  the 
perpendicular  bisector. 


148  PLANE  GEOMETRY 

Prove  the  following  by  Th.  86  and  Cor. 

Ex.  1.  The  diagonals  of  a  rhombus  or  of  a  square  bisect 
each  other  at  right  angles. 

Ex.  2.     One  diagonal  of  a  kite  bisects  the  other  at  right  angles. 

Ex.  3.  If  two  circles  intersect,  the  line  of  centers  is  a  perpen- 
dicular bisector  of  the  common  chord. 

Ex.  4.  If  two  equal  circles  intersect,  the  segment  joining  the 
centers  and  the  common  chord  bisect  each  other  at  right  angles. 

Ex.  5.  The  perpendicular  bisector  of  a  chord  passes  through 
the  center  of  the  circle. 

Ex.  6.  A  radius  to  the  mid-point  of  an  arc  is  the  perpen- 
dicular bisector  of  the  chord  of  the  arc. 

Ex.  7.  Two  tangents  to  a  circle  where  the  center  is  O  inter- 
sect at  point  X.  Prove  that  OX  is  the  perpendicular  bisector  of 
the  chord  joining  the  points  of  contact  of  the  two  tangents. 

OTHER   SIMPLE  LOCI 

183.  Find  the  loci  called  for  in  the  following  exercises; 
give  complete  proof  for  each  as  for  Ths.  85  and  86. 

Find  the  locus  of  a  point  which  is 

Ex.  1.     At  a  given  distance  from  a  given  point. 

Ex.  2.     At  a  given  distance  from  a  given  straight  line. 

Ex.  3.     Equally  distant  from  two  given  parallels. 

CONCURRENT  LINES 

183.  The  two  following  exercises  are  preliminary  and  may 
be  quoted  as  theorems  in  proving  Ths.  87  and  89. 

Ex.  1.  Lines  that  are  perpendicular  to  intersecting  straight 
lines  will  intersect. 

Suggestion.  Use  an  indirect  proof.  AD  and  DB  are 
the  given  intersecting  lines.  Join  A  and  B.  Show  that. 
ii  AC  and  BC  were  parallel,  Z1+  Z2  would  equal  two 
right  angles  (Fig.  266). 

Ex.  2.     The  bisectors   of   any  two   angles   of   a      Fig.  266 
triangle  will  intersect. 


LOCI  149 

184.  Theorem  87.  The  perpendicular  bisectors  of  the 
sides  of  a  triangle  are  concurrent  at  a  point  which  is  equally 
distant  from  the  vertices. 


Fig.  267 

Hypothesis:  AABC  is  any  triangle.  Lines  x,  y,  and  z 
are  the  perpendicular  bisectors  of  AB,  EC,  and  AC  respec- 
tively. 

Conclusion:  x,  y,  and  z  are  concurrent  at  a  point  equally 
distant  from  the  vertices. 

Analysis: 
I.  To  prove  x,  y,  and  z  concurrent,  prove 

a.  That  x  and  y  intersect  at  some  point,  as  0. 

b.  That  the  intersection  of  x  and  y  is  on  z. 

IL  To  prove  that  the  intersection  of  x  and  y  is  on  z, 
prove  that  0  is  equally  distant  from  A  and  C. 
III.   .'.  prove  0  equally  distant  from  A  and  B,  also  from 
B  and  C. 
Proof: 

STATEMENTS 

I.  X  and  y  intersect  at  some  point,  as  at  0. 
IL  a.  X  is  locus  of  points  equally  distant  from  A  and  B. 
b.    .'.  O  is  equally  distant  from  A  and  B. 

III.  a.  y  is  locus  of  points  equally  distant  from  B  and  C. 
b.    .:  0  is  equally  distant  from  B  and  C. 

IV.  .'.  O  is  equally  distant  from  A  and  C. 

'  V.  a.  z  is  locus  of  points  equally  distant  from  A  and  C. 
b.   .'.  O  is  on  z. 
Let  the  pupil  give  the  reasons. 
11 


150  PLANE  GEOMETRY 

Theorem  88.     The  altitudes  of  a  triangle  are  concurrent. 


Fig.  268 

Hypothesis:  AABC  is  any  triangle,  x,  y,  and  «  are  the 
altitudes  to  sides  AB,  EC,  and  CA  respectively. 

Conclusion:     x,  y,  and  z  are  concurrent. 

Analysis:  To  prove  x,  y,  and  z  conctirrent,  construct 
through  A,  B,  and  C  lines  parallel  to  the  opposite  sides  of 
AABC  and  prove  that  x,  y,  and  z  are  the  perpendicular 
bisectors  of  the  sides  of  the  triangle  so  formed. 

Let  the  pupil  put  in  the  construction,  complete  the  analysis,  and 
give  the  proof. 

185.  Theorem  89.  The  bisectors  of  the  angles  of  a  tri- 
angle are  concurrent  at  a  point  equally  distant  from  the 
sides  of  the  triangle. 


Hypothesis:     AABC  is  any  triangle,     x,  y,  and  z  bisect 
AA,  B,  and  C  respectively. 

Conclusion:     x,  y,  and  z  are  concurrent  at  a  point  equally 
distant  from  the  sides. 


LOCI  161 

Analysts: 

I.  To  prove  x,  y,  and  z  concurrent,  prove 

a.  That  x  and  y  intersect  at  some  point,  as  at  O. 

b.  That  the  intersection  of  x  and  y  is  on  z. 

11.  To  prove  that  the  intersection  of  x  and  y  is  on  z, 
prove  that  O  is  equally  distant  from  AC  and  BC. 
III.    .*.  prove  that  O  is  equally  distant  from  AC  and  AB, 
also  from  AB  and  ^C 

186.     Problem  12.    To  circumscribe  a  circle  about  a  tri- 
angle. 

Note.     This     problem    has    been     discussed    elsewhere     (§139). 
The  discussion  here  should  be  based  upon  Th.  87. 

Problem  13.    To  inscribe  a  circle  in  a  given  triangle. 


Fig.  270 

Given  AABC. 

To  inscribe  a  circle  in  AABC. 
Analysis  and  constmction: 
I.  To  construct  a  circle  inscribed  in  AABC,  construct 

a  circle  tangent  to  the  sides  of  AABC. 
II.    /.  find  a  point,  such  as  O,  so  situated  that  perpen- 
diculars from  O  to  the  sides,  such  as  OE,  OF,  and 
OG,  are  radii. 

III.  .'.  construct 

IV.  To  prove  the  perpendiculars  radii,  prove  them  equal. 
Let  the  pupil  give  the  directions  and  the  proof. 

Exercise.     Construct  a  circle  tangent  to  one  side  of  a  triangle 
and  to  two  sides  extended. 

A  circle  tangent  to  one  side  of  a  triangle  and  to  two  sides 
extended  is  said  to  be  escribed  to  the  triangle. 


152  PLANE  GEOMETRY 

DETERMINATION  OF  POINTS  BY  THE 
INTERSECTION  OF  LOCI 

187.  Give  analysis,  directions,  proof,  and  discussion  for 
the  following  exercises. 

Ex.  L     Find  in  a  given  line  a  point  that  is  equally  distant 
from  two  given  points. 
A  nalysis: 
I.  Let  /  represent  the  given  line  and  A  and  B  the  given  points. 
II.  To  find  a  point  in  Hne  /  equally  distant  from  A  and  B,  find  the 
intersection  of  line  /  with   the  locus  of  points  equally  distant 
from  A  and  B. 
Let  the  pupil  give  the  directions  and  the  proof. 
Discussion: 

How  many  points  can  be  found  as  required? 

Are  there  any  special  positions  of  the  given  line  or  the  given  points 
that  will  alter  the  results?     Give  all  reasons. 

Find  in  a  given  line  a  point  that  is 

Ex.  2.     At  a  given  distance  from  a  given  point. 

Ex.  3.     Equally  distant  from  two  given  parallel  lines. 

Ex.  4.     At  a  given  distance  from  a  second  given  line. 

Ex.  5.  Equally  distant  from  two  given  intersecting  lines. 
In  the  preceding  five  problems  the  required  point  must 
not  only  be  in  a  given  line,  but  must  also  fulfill  a  second 
requirement  which  calls  for  the  construction  of  a  locus.  In 
the  problems  that  follow,  the  construction  of  two  loci  are 
necessary  in  order  to  determine  the  point.  The  analysis 
should  state  clearly  what  loci  are  needed.  We  have  seen 
that  the  intersection  of  two  straight  lines  locates  a  point 
and  that  the  intersection  of  a  straight  line  and  a  circle 
locates  two  points.  As  some  loci  consist  of  two  straight 
lines,  it  is  often  possible  to  find  more  than  one  point  that 
fulfills  the  requirements.  The  pupil  should  begin  by  draw- 
ing a  figure  to  illustrate  the  maximum  number  of  points 
possible,  and  after  giving  the  analysis  with  the  directions 
for  the  construction  and  the  proof  he  should  discuss  and 
draw  figures  for  all  special  cases. 


LOCI  153 

Ex.  6.  Find  a  point  which  is  at  a  given  distance  from  a 
given  point  C  and  at  the  same  time  equally  distant  from 
two  given  points  A  and  B. 

A  nalysis: 
I.  To  find  all  points  that  are  at  a  given  distance  from  point  C, 

construct  the  locus  of 

II.  To  find  all  points  equally  distant  from  A  and  B,  construct  the 

locus  of 

III.    .'.  construct 

Let  the  pupil  give  definite  directions  for  the  construction,  also  the 
proof  and  the  discussion. 

Ex.  7.  Points  may  be  found  which  will  fulfill  any  two  of  the 
following  requirements: 

a.  Be  at  a  given  distance  (1>2  in.)  frbm  a  given  point. 
h.  Be  at  a  given  distance  (1>^  in.)  from  a  given  line. 

c.  Be  equally  distant  from  two  given  points. 

d.  Be  equally  distant  from  two  given  parallel  lines. 

e.  Be  equally  distant  from  two  given  intersecting  lines. 
State  and  solve  problems  made  by  combining  into  pairs  the 

requirements  given  above. 

LOCI   OF   CENTERS   OF   CIRCLES 
DETERMINATION   OF  THE  LOCI 

188.  In  the  determination  of  the  loci  of  centers  of  circles, 
we  may  think  of  a  circle  as  changing.     Fig.  271  shows  a 
number  of  circles  of  the  same  radius  tan- 
gent to  line  I.     A  great  many  more  such    (^^TT^T^^fHT^'T^ 


circles   might    be    drawn.     We   might, 

however,  consider  these  as  various  posi-  Fig.  27] 

tions  of  one  circle  which  rolls  or  slides 

along  the  line.     In  the  same  way  in  Fig. 

272  a  number  of  circles  are  shown  tangent 

to  AB  and  AC\  but   we  might  consider 

them  as  various   positions  of  one-  circle 

which  expands  and  contracts   as   it  rolls  Fig.  272 

or  slides  between  AB  and  AC. 


154  '  PLANE  GEOMETRY 

Find  the  locus  called  for  in  each  of  the  following  exercises 
and  give  complete  proof  for  each. 

Ex.  1.  ^Find  the  locus  of  the  centers  of  circles  which  are 
tangent  to  the  sides  of  an  angle. 

I.  To  find  the  locus,  sketch  in  a  number  of  circles  tangent  to  the 

sides  of  the  given  angle  Z  A .     Let  AX  be  the  locus. 
II.  To  prove  that  AX  is  the  required  locus,  prove  that 

a.  Every  point  in  AX  may  be  the  center  of  a  circle  tangent 

to  the  sides  of  the  angle,  AB  and  A  C,  and 

b.  The  center  of  every  circle  tangent  toAB  and  A  C  lies  in  AX. 
III.  Discussion.     AX  may  be  considered  as  the  path  of  a  circle  that 

•  expands  and  contracts  as  it  rolls  or  slides  between  the  rays 
AB  and  AC. 

Find  the  locus  of  the  center  of  a  circle  which 

Ex.  2.  Is  tangent  to  each  of  two  intersecting  lines. 

Ex.  3.  Passes  through  two  given  points. 

Ex.  4.  Is  tangent  to  a  given  line  at  a  given  point. 

Ex.  5.  Is  tangent  to  a  given  line  and  has  a  given  radius. 

Ex.  6.  Is  tangent  to  a  given  circle  at  a  given  point. 

Ex.  7.  Is  tangent  to  a  given  circle  and  has  a  given  radius. 

Ex.  8.  Passes  through  a  given  point  and  has  a  given  radius. 

CONSTRUCTION   OF   CIRCLES 

189.  Make  the  constructions  called  for  in  the  follov^^ing 
exercises.  The  center  of  the  required  circle  is  the  inter- 
section of  two  loci.     The  discussion  is  often  interesting. 

Ex.  1.  Construct  a  circle  that  shall  be  tangent  to  a  given 
line  at  a  given  point  and  pass  through  a  second  given  point. 

Ex.  2.  Construct  a  circle  tangent  to  the  sides  of  an  angle  and 
to  one  of  these  sides  at  a  given  point. 

Ex.  3.  Construct  a  circle  that  shall  be  tangent  to  a  given 
circle  at  a  given  point  and  pass  through  a  second  given  point. 

Construct  a  circle  of  given  radius  that  shall 
Ex.  4.     Be  tangent  to  each  of  two  given  intersecting  lines. 
Ex.  5.     Pass  through  a  given  point  and  be  tangent  to  a  given 
line  or  to  a  siven  circle. 


LOCI  155 

SUPPLEMENTARY  EXERCISES 
EXERCISES  INVOLVING  CONCURRENT  LINES 

190.  To  the  tests  for  concurrent  lines  given  in  this  chapter 
should  be  added  Th.  49,  given  in  chapter  iv. 

Each  triangle  has  four  sets  of  concurrent  lines.  The 
intersection  of  each  set  has  a  special  name  as  shown  below. 

I.  The  medians centroid  or  center  of  gravity 

II.  Perpendicular  bisectors  of  the  sides  . . . .  circumcenter 

III.  The  altitudes orthocenter 

IV.  Bisectors  of  the  angles incenter 

Note.  Be  prepared  to  prove  the  theorems  on  which  any  of  the 
following  exercises  depend.     Review  Th.  49. 

1.  The  centroid,  incenter,  circumcenter,  and  orthocenter  of 
an  equilateral  triangle  coincide.  What  facts  do  you  know  con- 
cerning this  point? 

2.  In  Fig.  273,  ABC  is  an  equilateral  triangle. 
AYf  BZy  and  CX  are  equal  distances  laid  off  on 
the  medians  AF^  BG,  and  CE  respectively.  Prove 
that  XYZ  is  an  equilateral  triangle.  '  "    Fig'' 273 

3.  If  on  the  sides  of  ZA  equal  distances  AB  and  AC  are 
laid  off,  perpendiculars  erected  to  the  sides  of  the  angle  at  B  and 
C  will  intersect  on  the  bisector  of   ZA. 

4.  The  bisector  of  two  exterior  angles  of  a  triangle  and  of  the 
opposite  interior  angle  are  concurrent. 

5.  The  medians  and  diagonals  of  a  square  are  concurrent. 

6.  Construct  A  ABC,  given  ^5  =  5.8  cm.,  the  median  from 
A  =  7.S  cm.,  and  the  median  from  J5  =  6.4  cm. 

Suggestion.     See  Th.  49. 

7.  Construct    A  ABC,   given   AC,  AB,  and 
the  median  from  A  (Fig.  274). 

8.  Construct     A  ABC,      given     the     three  /     ^. 
medians.                                                                             fy" 

Suggestion.     Reduce  this  exercise  to  the  preceding  ^ 

by  means  of  Th.  49.  Fig.  274 


156  PLANE  GEOMETRY 

EXERCISES  INVOLVING  CONSTRUCTION  OF  CIRCLE3 

191.  Notice  (1)  that  a  circle  can  be  circumscribed  about 
a  polygon  if  there  is  a  point  equally  distant  from  the  ver- 
tices, that  is,  if  the  perpendicular  bisectors  of  the  sides  are 
concurrent,  and  (2)  that  a  circle  can  be  inscribed  in  a  poly- 
gon if  there  is  a  point  equally  distant  from  the  sides,  that 
is,  if  the  bisectors  of  the  angles  are  concurrent. 

1.  Construct  a  circle  of  given  radius  that  shall  pass  through 
two  given  points. 

2.  Given  the  base  of  an  isosceles  triangle  and  the  radius  of 
the  circumscribed  circle,  to  construct  the  triangle. 

3.  Can  a  circle  be  passed  through  four  arbitrary  points? 
Why? 

4.  When  will  the  perpendicular  bisectors  of  the  sides  of  a 
triangle  meet  on  one  of  the  sides  of  the  triangle?  When  will  they 
meet  within  the  triangle?  When,  will  they  meet  without  the 
triangle?    Why? 

5.  Can  a  circle  be  circumscribed  about  a  parallelogram? 
a.  rectangle?  In  case  the  circumscribed  circle  is  possible,  give 
analysis,  directions,  and  proof.  In  case  the  circumscribed  circle 
is  not  possible,  show  why. 

6.  Answer  the  questions  in  Ex.  5  for  an  isosceles  trapezoid 
and  for  a  trapezoid. 

7.  How  many  circles  can  be  constructed  tangent  to  each  of 
three  intersecting  lines?     Discuss  all  possible  cases. 

8.  Can  a  circle  be  constructed  tangent  to  each  of  three  arbi- 
trary lines?     Discuss  all  possible  cases. 

9.  Inscribe  a  circle  in  a  given  rhombus. 

10.  Inscribe  a  circle  in  a  given  kite. 

11.  Circumscribe  about  a  given  circle  an  isosceles 
right  triangle. 

12.  Construct  in  full  Fig.  275. 

Fig.  275 

13.  Prove  that  the  center  of  the  circle  circumscribed  about  an 
equilateral  triangle  is  also  the  center  of  the  inscribed  circle. 


LOCI 


157 


B 

Fig.  276 


/^'    \  V.J 


14.  Construct  the  inscribed,  the  circumscribed,  and  the  three 
escribed  circles  of  an  equilateral  triangle.  Prove  that  the  radius 
of  the  inscribed  circle  is  ^4  the  radius  of  the  circumscribed  circle 
and  }4  the  radius  of  the  escribed  circles. 

15.  Make  the  complete  drawing  for  the 
molding  shown  in  Fig.  276.  The  arc  ^40  is  ^t 
tangent  to  line  AX  at  A.  The  arc  BO  is 
tangent  to  line  BY  at  B.  Both  arcs  pass 
through  O,  the  mid-point  of  segment  AB. 
Are  the  arcs  AO  and  BO  tangent  to  each  other? 

Note.     Compound  curves  may  be  made  from      

tangent  circles  as  shown  in  Fig.  277.  These  may 
be  used  for  moldings  as  shown  in  Fig.  276,  or  for 
other  architectural  details.  p       277 

16.  Show  by  a  complete  drawing  how  to  construct  Fig.  278- 
Two  diagonal  streets  meet  at  A.  The  corner  building 
has  a  front  that  is  an  arc  of  a  circle  of  a  given  radius 
tangent  to  each  of  the  streets.  If  Z  /I  is  60°  and  the 
radius  of  the  circle  50  ft.,  what  will  be  the  length  of 
AB  and  BC  d  B  and  C  are  the  points  of  tangency? 

A  sector  of  a  circle  is  a  figure  bounded  by  two  radii  and 
the  subtended  arc. 

17.  Construct  a  circle  inscribed  in  a  sector  of  a  given  circle. 

c  B 

18.  Fig.   279   shows   a   decorated  rafter  design.   ^ 

Z  A  CB  is  a  right  angle.     A  DB  is  tangent  to  the  sides 

of   ZC.     Construct  the  small  circle  tangent  to  the 

sides  of  ZC  and  to  ADB.  ^ 

Fig.  279 

19.  Fig.  280  shows  a  decorated  tile  design.    ABCD\ 
is  a  square  with  its  diagonals.     Construct  (D  O  and 
O'  inscribed    in  the  A  A  DC   and  ABC  respectively. 
Construct  the  quadrants  with  A  and  C   as  centers 
tangent  to  (D  0  and  0'.  p^^    280 

Note.  The  possibilities  of  this  figure  as  a  design  unit  may  be  seen 
by  drawing  several  figures  of  the  same  size,  like  Fig.  280,  and  placing 
them  together  in  various  positions. 


Fig.  278 


158 


PLANE  GEOMETRY 


20.  Fig.    281    shows    a    window    and    rafter 
is  an  isosceles  right  triangle.    Construct  O  O 
inscribed  in  A  ABC  and  (D  X  and  Y  tangent  to 
the  sides  of  the  triangle  and  to  0  0. 


ABC 


Fig.  281 


21.  Inscribe  a  trefoil  in  a  given  circle  (Fig.  282). 

Note.     The  three  small  circles  are  tangent  to  the 
large  circle  and  to  each  other.  ■ 


22.  Show  how  to  construct  circles  which 
are  tangent  to  each  of  two  concentric  circles. 

23.  Any  point  in  the  perpendicular  bisec- 
tor of  the  segment  joining  the  centers  of  two 
equal  circles  may  be  used  as  the .  center  of  a 
circle  tangent  to  each  of  the  two  given  circles. 
Use  various  positions  of  the  two  given  circles. 

Note.  The  perpendicular  bisector  mentioned 
in  Ex.  23  is  a  part  of  the  locus  of  the  center  of 
a  circle  tangent  to  each  of  two  equal  circles.  The 
remainder  of  the  locus  is  beyond  the  province 
of  elementary  geometry,  but  may  be  readily 
found  by  experiment. 

24.  In  Fig.  283,  DF  and  BC  are  concen- 
tric with  point  A  as  center.  DE  and  A  C  are 
concentric  with  point  B  as  center.  DF  and 
DE  are  tangent  at  D.  AD  =  DB.  AF  and 
BE  are  drawn  with  D  as  center  and  J/2  AB 
as  radius.  Construct  O  0  tangent  to  AC, 
BC,  FD,  and  DE. 

_25.  In  Fig.  284,  A  ABC  is  equilateral.  AB, 
BC,  and  AC  are  drawn  with  AB  as  radius  and 
C,  A ,  and  B  as  centers  respectively.  Construct 
0  O  tangent  to  AB,  BC,  and  AC. 

Note.     Fig.  284  is  from  a  church  window  design. 


Fig.  282 


Fig.  283  a 


Fig.  284 


LOCI  159 

MISCELLANEOUS  EXERCISES 

192.  1.  What  is  the  locus  of  the  mid-points  of  all  equal  chords 
of  a  circle? 

2.  What  is  the  locus  of  the  mid-points  of  a  series  of  parallel 
chords? 

3.  What  is  the  locus  of  the  mid-points  of  segments  drawn  from 
a  given  point  to  a  given  line? 

t4.  Find  the  locus  of  the  vertices  of  triangles  having  a 
given  base  and  a  given  vertex  angle.  What  is  this  locus  if  the 
given  angle  is  a  right  angle? 

Construct  a  triangle,  given 

5.  The  base,  the  vertex  angle,  and  the  median  from  the  ver- 
tex to  the  base. 

6.  The  base,  the  vertex  angle,  and  the  altitude. 

7.  The  base,  the  vertex  angle,  and  one  base  angle. 

8.  The  base,  the  vertex  angle,  and  one  side. 

9.  Find  the  locus  of  the  mid-points  of  segments  drawn  to  the 
circle  from  a  fixed  point  (a)  without  the  circle,  (b)  on  the  circle, 
(c)  within  the  circle. 

10.  A  circle  of  radius  5  inches  contains  a  moving  chord  AB, 
length  8  inches,  which  is  divided  into  four  equal  parts  by  the 
points  P,  Q,  R.  Determine  the  loci  of  P,  Q,  and  R. — College  En- 
trance Examination  Board,  Plane  Geometry   Examination,  1913. 

11.  A  series  of  parallelograms  stand  on  the  same  base  and 
between  the  same  parallels.  Find  the  locus  of  the  intersection 
of  the  diagonals. 

12.  From  any  point  in  the  base  of  a  triangle  straight  lines 
are  drawn  parallel  to  the  sides.  Find  the  locus  of  the  intersection 
of  the  diagonals  of  all  the  parallelograms  that  can  be  thus  formed. 

13.  Find  the  locus  of  the  points  at  which  two  equal  segments 
of  a  straight  line  subtend  equal  angles. 


160  PLANE   GEOMETRY 

14.  Find  the  locus  of  the  extremities  of  tangents  to  a  circle 
that  have  the  same  length. 

15.  Find  the  locus  of  points  from  which  tangents  to  a  given 
circle  meet  at  a  given  angle. 

16.  Find  the  locus  of  points  of  contact  of 
tangents  drawn  from  a  fixed  point  to  a  system 
of  concentric  circles. 

17.  Construct  a  series  of  circles  tangent  to 
each  other  at  the  same  point  (Fig.  285). 
Find  the  locus  of  points  of  contact  of  tangents 
drawn  to  these  circles  from  any  point  in  the 
common  tangent.  Fig.  285 

18.  Two  circles  are  tangent  to  a  given  straight  line  at  two  given 
points  and  are  also  tangent  to  each  other.     Find 
the  locus  of  points  of  tangency  of  the  two  circles. 

Suggestion.  Let  A  and  B  he  the  two  given 
points,  C  the  point  of  tangency  of  the  two  circles. 
Prove  that  CO  is  always  equal  to  14  A  B  (Fig.  286). 

19.  In  Fig.  287,  AACB  is  a  right  triangle  with 
the  right  angle  at  C.  BCDE  is  the  square  con- 
structed on  side  BC.  Find  the  locus  of  the  vertex 
D  asC  moves  about  the  semicircle  BCA .    Use  Ex .  4 .       Fig.  287 

20.  Upon  a  line  segment  AB  an  arc  of  a  circle  containing 
240°  is  constructed  and  in  the  arc  any  chord  CD  having  an  arc 
of  60°  is  drawn.  Find  the  locus  (a)  of  the  point  of  intersec- 
tion oi  AC  and  BD,  (b)  of  the  point  of  intersection  oi  AD  and  BC. 
— College  Entrance  Examination  Board,  Plane  Geometry  Exami- 
nation, 1906. 

21.  Let  A  and  B  be  two  fixed  points  on  a  given  circle  and  P 
and  Q  the  extremities  of  a  variable  diameter  of  the  same  circle. 
Find  the  locus  of  the  point  of  intersection  of  the  straight  lines  AP 
and  BQ. — College  Entrance  Examination  Board,  Plane  Geometry 
Examination,  1908.  , 

22.  From  a  given  point  on  a  circle  draw  the  chords  that  are 
bisected  by  a  given,  chord.  Is  it  always  possible  to  draw  such 
chords?  Give  reasons  for  your  answer. — College  Entrance  Exami- 
nation Board,  Plane  Geometry  Examination,  1907. 


CHAPTER  IX 

Ratio  and  Proportion 

MEASUREMENT  OF  SEGMENTS 

193.  To  measure  a  segilient  is  to  find  the  number  of  times 
that  it  contains  another  segment  which  is  taken  as  a  unit. 

In  measuring  the  segment  two  methods  are  possible.  By  the  first 
method  the  unit  is  actually  laid  down  successively  on  the  segment  to 
be  measured.  This  method  might  be  used  in  finding  the  length 
of  a  room  if  nothing  but  a  yardstick  were  at  hand.  By  the  second 
method  another  segment,  upon  which  the  unit  and  its  subdivisions 
are  already  marked,  is  laid  beside  the  segment  to  be  measured. 

The  number  found  is  called  the  measure  number,  the 
measure,  or  the  length  of  the  segment. 

194.  A  segment  is  said  to  be  measured  exactly  if  it  will 
contain  the  unit  without  remainder. 

A  segment  that  is  not  measured  exactly  may  be  measured 
approximately. 

In  considering  the  theoretical  measurement  of  segments 
we  have  the  following  cases: 

First:  The  unit  chosen  may  be  contained  in  the  given 
segment  without  remainder.  The  length  of  the  segment  is 
then  an  integer.     The  segment  is  measured  exactly. 

Illustration  1.  If  the  unit  chosen  is  a  segment  of  one  inch,  it  may 
be  contained  in  a  given  segment  5  times  with  no  remainder.  The 
length  of  the  given  segment  is  5. 

Second:  The  unit  chosen  may  not  be  contained  in  the 
given  segment  without  remainder.  In  such  cases  it  may 
happen  that  some  fraction  of  the  unit  can  be  found  that 
will  measure  the  segment  exactly.     If  one  inch  is  the  unit 

161 


162  PLANE  GEOMETRY 

chosen,  3^  in.,  J^  in.,  34  in.,  or  .1  in.  may  be  used.  If  such 
a  unit  can  be  found,  the  segment  is  said  to  be  measured 
exactly,  tn  such  a  case  the  length  will  be  an  integer  when 
expressed  in  terms  of  the  new  unit,  but  a  fraction  when 
expressed  in  terms  of  the  old  unit.  In  either  case  the  seg- 
ment has  been  measured  exactly. 

Illustration  2.  If  the  unit  chosen  is  one  inch,  it  may  be  contained 
in  a  given  segment  7  times  with  a  remainder  less  than  one  inch;  but 
when  3^  inch  is  chosen  as  a  unit  the  measure  may  come  out  exactly  31. 
The  length  is  31  quarter-inches,  or  7%  in.;  or 

Illustration  3.  The  unit  chosen,  one  inch,  may  be  contained  in  a 
given  segment  3  times  with  a  remainder  less  than  one  inch;  a  smaller 
unit,  .1  in,,  maybe  contained  in  the  segment  32  times  with  a  remainder 
less  than  .1  in.;  a  still  smaller  unit,  .01  in.,  may  be  contained  in  the 
segment  324  times  with  a  remainder  less  than  .01  in.;  but  the  unit 
.001  in.  may  be  contained  exactly  3247  times.  The  length  of  the  seg- 
ment is  3247  thousandths  of  an  inch,  or,  as  it  is  usually  written,  3.247  in. 

Third:  It  may  happen  that  no  subdivision  of  the  unit 
can  be  found  that  will  measure  the  segment  exactly.  In 
such  a  case  we  may  obtain  an  approximate  measure.  By 
subdividing  the  unit  used  the  approximation  may  be  made 
as  close  as  desired. 

Illustration  4.  The  unit  chosen,  one  inch,  may  be  contained  in  the 
given  segment  5  times  with  a  remainder  less  than  one  inch.  In  this 
case  5  would  be  an  approximate  length  of  the  segment.  If  we  choose  .1 
in.  as  a  unit,  it  may  be  contained  in  the  given  segment  56  times  with  a 
remainder  less  than  .1  in.  We  now  have  5.6  in.  as  an  approximate 
length.  But  if  we  should  choose  .01  in.  as  a  unit,  we  could  get  a  still 
closer  approximation.  It  might  happen  that  .01  in.  would  be  contained 
in  the  segment  562  times  with  a  remainder  less  than  .01  in.  The 
approximate  length  is  now  5.62  in.  This  process  might  be  continued 
indefinitely. 

195.  In  actual  practice  an  exact  measurement  can  never 
be  obtained.  We  cannot  be  sure  that  a  segment  is  exactly 
7  in.  or  7J4  in.  long.  In  trying  to  measure  the  segment,  the 
end  will  fall  between  two  marks  on  the  scale.  Either  of 
these  gives  an  approximation  to  the  length  of  the  segment, 
one  a  little  too  small  and  one  a  little  too  large.     It  is  the 


RATIO  AND  PROPORTION  '  163 

usual  practice  to  use  the  nearest  one  as  the  approximate 
length  of  the  segment.  Often  extremely  close  approxima- 
tions are  necessary,  but  the  degree  of  accuracy  depends 
upon  the  fineness  of  the  scale  used  and  the  definiteness  of 
the  end  of  the  segment  to  be  measured. 

Ex.  1 .  Draw  a  segment  3)4  in.  long.  Measure  it  in  centimeters 
and  millimeters.  Make  two  approximate  measures,  one  as  close 
as  possible  but  a  little  too  small,  the  other  as  close  as  possible  but 
a  little  too  large.  From  each  result  compute  the  number  of  centi- 
meters to  an  inch  and  the  number  of  inches  to  a  centimeter.  Com- 
pare your  result  with  the  government  standard  equivalent  (p .  300) . 

Ex.  2.  Draw  a  segment  5.6.cm.  long.  Make  two  approximate 
measures  of  this  segment  in  inches  and  sixteenths  of  an  inch. 
From  your  results  make  the  computations  called  for  in  Ex.  1 . 

RATIOS 
DEFINITIONS 

196.  The  ratio  of  two  numbers  is  the  relation  expressed 
by  dividing  one  number  by  the  other.  The  ratio  of  a  to  6 
is  written  in  two  forms — a  :b  or|-.  It  is  read  the  ratio  of 
a  to  b.  Two  numbers  are  involved,  the  first  term  or  divi- 
dend, also  called*  the  antecedent,  and  the  second  term  or 
divisor,  also  called  the  consequent. 

When  we  say  that  the  ratio  of  12  to  4  is  3,  or  ^  =  3,  we 
mean  that  12  is  3  times  4;  when  we  say  that  the  ratio  of  a 
to  b  is  r,  or  -J  =r,  we  mean  that  a  is  r  times  6,  or  a  =  br. 

The  quotient  r  is  sometimes  called  the  value  of  the  ratio. 
It  is  the  common  practice  to  use  the  term  ratio  to  mean 
either  the  indicated  relation -f  or  the  quotient  r;  either  ^ 
or  3.  In  any  case  the  two  numbers  or  terms  a  and  b  are 
always  involved. 

Two  ratios  that  have  the  same  value  are  said  to  be  equal. 
Exercise.  Express  the  following  ratios  in  decimals  of  three  places : 
5^       n       45       12      ^       2       _3 
9'     27'     59'     25'      19'      24'     44' 


164  PLANE   GEOMETRY 

RATIO    OF   SEGMENTS 

197.  By  the  ratio  of  two  segments  is  meant  the  ratio  of 
their  measures  when  expressed  in  the  same  unit. 

Exercise.  Draw  two  segments,  one  2  cm.  and  one  3  cm.  long. 
Measure  each  in  inches  and  sixteenths  of  an  inch  or  in  inches 
and  tenths  of  an  inch.     Find  an  approximate  ratio. 

198.  Two  segments  are  said  to  be  commensurable  if  they 
can  be  measured  exactly  by  a  common  unit  of  measure. 

Two  segments  are  said  to  be  incommensurable  if  there  is 
no  common  unit  that  will  measure  each  exactly.  We  shall 
later  prove  that  the  side  and  the  diagonal  of  a  square  are 
incommensurable.  If  the  side  of  a  square  is  one  inch,  the 
diagonal  is  V2~ inches,  an  irrational  number. 

An  irrational  number  is  a  number  that  cannot  be  expressed 
as  an  integer  or  as  the  quotient  of  two  integers. 

The  ratio  of  two  commensurable  segments  is  an  integer 
or  a  fraction. 

The  ratio  of  two  incommensurable  segments  is  an  irra- 
tional number. 

Other  illustrations  of  incommensurables  will  be  met  later. 
Two  especially  may  be  mentioned: 

1.  The  side  of  an  equilateral  triangle  and  the  altitude  of 
the  same  triangle  are  incommensurable. 

2.  The  diameter  of  a  circle  is  incommensurable  with  the 
circumference  of  the  circle. 

Note.  The  nature  of  the  decimals  that  correspond  to  rational  and 
to  irrational  numbers  is  interesting  and  should  be  noted.  Fractions 
are  rational  and  when  reduced  to  decimals  give  decimals  that  either 
terminate  or  repeat,  for  example: 

He  =  -0625  >^  =  .333+ or  .3 

K  =  .1666+  or  .16  H  =  .142857142857  or  .1-12857 

Irrational  numbers,  on  the  other  hand,  give  decimals  which  neither 
terminate  nor  repeat.  An  inexact  root  like  V2~or  V  3" is  an  irrational 
number,  but  not  the  only  kind  of  an  irrational  number.  Another 
example  is  the  number  called  T  (pi)  (see  §§  298,  301,  and  311). 


RATIO  AND  PROPORTION  165 

ASSUMPTIONS  INVOLVING  RATIOS 

199.  The  following  assumptions  will  be  used;  As.  56 
expresses  the  fundamental  characteristic  of  ratios: 

As.  56.     Multiplying  or  dividing  both  terms  of  a  ratio  by 
the  same  number  does  not  change  the  value  of  the  ratio. 
As.  57.     Ratios  equal  to  the  same  ratio  are  equal. 
As.  58.     Equal  ratios  may  be  substituted  for  equal  ratios. 

THEORY   OF  PROPORTION 

DEFINITIONS  ^ 

200.  A  proportion  is  an  equality  of  ratios;  that  is,  if  two 
ratios  are  equal,  the  four  numbers  involved  are  in  proportion. 
A  proportion  may  be  written  in  two  forms,  a:b  =  c:d  or 

r-  =  t'  and  is  read,  a  is  to  6  as  <;  is  to  d,  or  the  ratio  a  to  6 
0     a 

equals  the  ratio  c  to  d.     The  extremes  of  the  proportion  are 

a  and  d.     The  means  are  b  and  c. 

Since  in  dealing  with  ratios  we  are  dealing  with  numbers, 

the  laws  of  algebraic  equations  apply  to  proportions. 

Exercise.     Find  the  value  of  x  in  each  of  the  following: 

^'    7  ~  X  55~64  ""•     6  2 

FUNDAMENTAL   THEOREMS   OF  PROPORTION 

201.  Theorem  90.  If  four  numbers  are  in  proportion, 
the  product  of  the  means  is  equal  to  the  product  of  the 
extremes. 

The  proof  is  left  to  the  pupil. 

Theorem  91.  If  the  product  of  two  numbers  equals  the 
product  of  two  other  numbers,  either  pair  of  factors  may  be 
made  the  extremes  and  the  other  pair  the  means  of  a 
proportion. 

Hypothesis:    ay  =  bx.        Conclusion:   -r  = 
Suggestion.     Divide  both  sides  of  ay  =  bx  by  by. 
12 


a       X 


166  PLANE  GEOMETRY 

Ex.1.    Given  ay  =  bx,prove-  =  -     -7  =  -»and  -^  =  -. 

X      y      0      a  X     a 

Ex.  2.  Derive  at  least  two  proportions  from  each  of  the 
following  equations: 

a.  ab  =  xy  c.  {x-\-y)  (x^y)  =  ab 

b.  aia+b)=:x(x+y)  d.  {a-1)  {x+l)  =  {a-{-l)  (x-l) 

Theorem  92.  If  three  terms  of  one  proportion  are  equal 
respectively  to  three  corresponding  terms- of  another  pro- 
portion, the  fourth  terms  are  equal. 

Hypothesis:     -t  =  -  and -r  =  - >     Conclusion:     x  =  y. 
The  proof  is  left  to  the  pupil. 

TRANSFORMATIONS   OF  PROPORTIONS 

202.  Theorem  93.  If  four  numbers  are  in  proportion, 
the  first  is  to  the  third  as  the  second  is  to  the  fourth;  that 
is,  they  are  in  proportion  by  mean  alternation. 

Hypothesis:    —  =  — •        Conclusion:    —  =  —  . 
by  X     y 

Suggestion.     First  prove  that  ay  =  bx;  then  use  Th.  91. 

Theorem  94.  If  four  numbers  are  in  proportion,  the 
fourth  is  to  the  second  as  the  third  is  to  the  first;  that  is, 
they  are  in  proportion  by  extreme  alternation. 

Hypothesis:    —  =  —-•       Conclusion:    ■^  =  —  ' 
by  b      a 

Let  the  pupil  give  the  proof. 

Theorem  95.  If  four  numbers  are  in  proportion,  the 
second  is  to  the  first  as  the  fourth  is  to  the  third;  that  is, 
they  are  in  proportion  by  inversion. 

Hypothesis:    —  =  — -       Conclusion:     _-=-2_. 
by  ax 

Let  the  pupil  give  the  proof. 

Ex.  1.     Verify  proportion   by  mean  and  extreme  alternation 

..      8      28     15      9  ,    a2     ac 

and  mversion  ^ri  j^  =  ^^  ^  =  ^,  and  -^  =  y^^ 


RATIO  AND  PROPORTION  167 

Theorem  96.  If  four  numbers  are  in  proportion,  the 
first  plus  the  second  is  to  the  second  as  the  third  plus  the 
fourth  is  to  the  fourth;  that  is,  they  are  in  proportion  by- 
addition.  This  is  sometimes  called  proportion  by  compo- 
sition. 


Hypothesis:    —  =— •        Conclusion: 

a+b 

x+y 

y 

rooj: 

STATEMENTS 

b     y 

2.  ay==bx. 

3.  ay-\-by  =  bx-\-by. 

4.  y{a+b)=b{x+y). 
f,    a-j-b    x-^y 

b          y 

Let  the  pupil  give  all  reasons. 

Theorem  97.  If  four  numbers  are  in  proportion,  the 
first  minus  the  second  is  to  the  second  as  the  third  minus 
the  fourth  is  to  the  fourth ;  that  is,  they  are  in  proportion  by 
subtraction.    This  is  sometimes  called  proportion  by  division. 

7     .        ax         ^      .     .          a—b     x—y 
Hypothesis:     -r-  =  —       Conclusion:     —r-  = • 

Ex.  2.    Verify  proportion  by  alternation,  inversion,  addition, 
and  subtraction  by  means  of  the  following: 

]^-?L     ?9_1)?     ^_^     ^-^ 
32~48'   35~28'    by~ bx'   by~ ab' 

Q,  X 

Ex.3.    If   -T  =  ~'  make  the  following  transformations: 

a.  First  by  mean  alternation  and  that  result  by  addition. 

b.  First  by  extreme  alternation  and  that  result  by  subtraction. 

c.  First  by  inversion  and  that  result  by  addition. 

d.  First  by  addition  and  that  result  by  extreme  alternation. 

e.  First  by  subtraction  and  that  result  by  inversion. 


168 


PLANE  GEOMETRY 


Ex.  4.     li  -r  =  — ,  prove  each  of  the  following: 


a. 


b. 


a-\-b 

_x+y 

a 

X 

a 
a+b 

X 

x-\-y 

a-^b 

x-y 

d. 


e. 


/. 


a±b_b_ 

X-\-y~  y 

a-{-b _  a 
x+y~  X 
a  —  b  a 
x—y      X 


a  —  b      b 
^'  x-y~J 

L    ^+^     x-\-y 
ti,  ,  — 

a—b     x~y 

'    —-^ 
^'  b^  ~'y^ 


RATIOS  OF  SEGMENTS  MADE  BY  PARALLELS 

FUNDAMENTAL   THEOREM 

203.  Theorem  98.  If  three  parallels  cut  two  transversals, 
the  segments  on  one  transversal  have  the  same  ratio  as  the 
corresponding  segments  on  the  other  transversal. 


Fig.  288 

Hypothesis:    AX\\BY\\CZ. 

AB  XY 

Conclusion:     The  ratio  of  -g^  =  the  ratio  of  y^  • 

Case  A:     When  AB  and  BC  are  commensurable. 

Analysis  and  construction: 

J    rr.      .         ,.         ,.    AB         ^  .     XY 

i.  io  snow  the  ratio  -^  =  the  ratio  y^»  use  the  meas- 
ures of  the  segments  AB,  BC,  XY,  and  YZ. 
II.  The  measures  of  AB  and  BC  are  to  be  assumed. 
III.  To  find  the  measures  of  XY  and  YZ,  divide  AB  and 
BC  into  segments  equal  to  their  common  unit  of 
measure  and  draw  lines  through  the  points  of 
division  parallel  to  CZ,  dividing  XY  and  YZ  into 
segments. 


RATIO   AND   PROPORTION  169 

Verification: 
I.  a.  Let  the  common  unit  of  measure  of  segments 
AB  and  BC  be  segment  p. 

b.  Let  the  measure  of  AB  =  m.     (In  the  figure  m  =  3.) 

c.  Let  the  measure  of  BC  =  n.     (In  the  figure  w  =  4.) 

d.  The  ratio  of  f77^=  — *     (In  the  figure  —  =  -t'^ 

BC      n      ^  *=*        w     4 

II.  a.  The  Hnes  parallel  to  CZ  divide  XY  into  m  equal 
segments  and  YZ  into  n  equal  segments  (Th.  45). 

b.  One  of  these  segments  may  be  taken  as  the  unit 

of  measure  oi  XY  and  YZ. 

^^         .       .XY    m 

c.  The  ratio  of  -rr^=—' 

YZ      n 

III.   .*.  the  ratio  of  -^^  =  the  ratio  of  y-y  • 

Note  1.     For  I  d  and  II  c  see  §  197,  the  ratio  of  two  segments. 

Note  2.  Since  we  assume  that  the  number  of  divisions  on  X  F  and 
YZ  are  respectively  equal  to  the  number  of  divisions  on  AB  and  BC, 
the  formal  reasoning  above  has  been  called  a  verification  rather  than  a 
proof. 

Case  B:  When  AB  and  BC  are  incommensurable.  Since 
it  is  not  possible  in  this  case  to  express  the  lengths  oi  AB 
and  BC  in  integral  or  fractional  terms  of  the  same  unit, 
the  argument  given  for  Case  A  cannot  be  used.  Case 
B  will  be  assumed  without  proof.  The  proof  is  possible, 
but  too  difficult  for  this  course. 

In  the  next  six  exercises  the  letters  refer  to  Fig.  288. 

Ex.1.  AB  =  7,        BC=9,      XF  =  17K,  find  FZ. 

Ex.2.  AB  =  12,     XF=15,     FZ=  18,  find  5C. 

Ex.3.  AB  =  S,       BC=yl2,XY  =  5,  find  YZ. 

Ex.4.  AB  =  3}^,   BC=IK    YZ  =  7,  find  XF. 

Ex.5.  AB  =  2^J^,XY  =  5,       FZ  =  7,  find  5C. 

Ex.  G.  BC=2.3,   XF  =  5.7,    FZ  =  9,  find  .45. 


170  PLANE  GEOMETRY 

APPLICATION   OF  THEOREM   98   TO   TRIANGLES 

204.  Theorem  99.  If  a  line  is  parallel  to  the  base  of  a 
triangle,  the  ratio  of  the  segments  on  one  side  equals  the 
ratio  of  the  corresponding  segments  on  the  other  side. 

Cor.  If  a  line  is  parallel  to  the  base  of  a  triangle,  one 
side  is  to  either  of  its  segments  as  the  other  side  is  to  its 
corresponding  segment. 

Suggestion.     Use  Th.  96. 
Ex.  1.     In  Fig.  289  prove  that 

^=^   £A-^    9A-9R 
CE  be'  cb~eb'  cb~ce' 

Ex.  2.     Prove  Th.  99  if  the  parallel  cuts  the 

Fig    289 
sides  of  the  triangle  extended. 

Ex.  3.  Verify  Th.  98  and  Th.  99  and  its  cor-,  by  measuring 
each  of  the  segments  in  Figs.  288  and  289  in  inches  and  sixteenths 
(or  tenths)  of  an  inch,  also  in  centimeters  and  millimeters,  and 
finding  the  ratios  from  the  measurements. 

In  the  next  eight  exercises  the  letters  refer  to  Fig.  289. 
Ex.4.     CD  =  S}^,    DA  =  3H,     CE  =  2}4,    find  £5. 


Ex.5. 

DA=2y2, 

CA  =  12}4, 

CB  =  9H, 

find  CE  and  EB. 

Ex.6. 

CD  =  m, 

CA=25, 

CB  =  15, 

find  CE  and  EB. 

Ex.  7. 

Z>^.=  3.6, 

EB  =  2A, 

CB  =  10.S, 

find  CD. 

Ex.8. 

CD  =  2^j^; 

DA  =3, 

CE  =  5, 

find  EB. 

Ex.9. 

DA=Q, 

CE  =  7, 

EB  =  7, 

find  CA, 

Ex.  10. 

CD  =  28, 

DA  =  14, 

CB  =  9S, 

find  CE  and  EB. 

Ex.  11. 

DA=2^2', 

CA=8, 

EB  =  15, 

find  EC. 

Ex.  12.     Name    the  pairs  of   segments    having  equal  ratios 
when  a  line  parallel  to  the  bases  of  a  trapezoid     ^^ 
cuts  the  non-parallel  sides  and  the   diagonals. 

Ex.  13.     Show  how  Th.  99   may  be  used  to  ^^        ^^  ^ 

find  an  inaccessible  distance  (see  Fig.  290).  Fig.  290 

Ex.  14.     Draw  a  figure  showing  four  parallels  cutting  two 
transversals.     Name  the  equal  ratios  formed. 


RATIO  AND  PROPORTION  171 

EQUAL  RATIO   TEST  FOR  PARALLELS 

205.  Theorem  100.  If  a  line  divides  the  sides  of  a  tri- 
angle so  that  one  side  is  to  one  segment  as  a  second  side 
is  to  its  corresponding  segment,  the  line  is  parallel  to  the 
third  side  of  the  triangle. 


Fig.  291 

Hypothesis:     AABC  is  any  triangle  with  DE  so  drawn 
CA     CB 

Conclusion:    DE  \\  AB. 
Analysis  and  construction: 
I.  To  prove  DE  \\  AB,  prove  that  DE  coincides  with  a 

line  II  AB, 
II.   .*.  construct  DX  from  D  ||  AB  and  prove  that  DE 
coincides  with  DX. 

III.  To  prove  that  DE  coincides  with  DX,  prove  that 

E  falls  on  X. 

IV.  To  prove  that  E  falls  on  X,  prove  that  CE  =  CX. 

M    rr.  ,u  .rT7     r^    u       CA_CB         CA  _CB 

V.  To  prove  that  CE  =  CX,  show  t^  -  ^  and  Yy)     'rx' 

.     The  proof  is  left  to  the  pupil.     For  V  use  Th.  92. 

Cor.  If  a  Hne  divides  the  sides  of  a  triangle  so  that  the 
ratio  of  the  segments  on  one  side  is  equal  to  the  ratio  of  the 
segments  on  the  other,  the  line  is  parallel  to  the  third  side 
of  the  triangle. 

Exercise.  If  a  line  divides  the  non-parallel  sides  of  a  trapezoid 
into  segments  having  the  same  ratio,  will  the  line  be  parallel  to  the 
base?    Give  proof, 


172  PLANE   GEOMETRY 

CONSTRUCTION    OF   PROPORTIONAL   SEGMENTS 
206.  Ex.  1.     By  algebra  divide  120  into  parts  that  shall  be  in 
the  ratio  of  7:8. 

Ex.  2.     Using  Prob.  7,  §  111,  find  ^f  of  a  given  segment. 
Ex.  3.     By  algebra  divide  144  into  three  parts  that  shall  be 
in  the  ratio  of  4,  5,  and  9. 

Problem  14.    To  divide  a  segment  into  two  segments 
that  shall  be  in  the  same  ratio  as  two  given  segments. 


Fig.  292 

Suggestion.     If  XY  is  the  given  segment,  show  how  to  construct 

h     k 
the  figure  so  that  you  can  prove  that  —  =  t"  '    Segments  lettered  alike 

are  equal.     Give  proof. 

Problem  15.    To  divide  a  given  segment  into  segments 
proportional  to  any  number  of  given  segments. 


Y^ »  ;i n Y 


Z' 

Fig.  293 


Suggestion.     It  is  necessary  to  construct    the   figure  so  that   you 

..    .h      k     n 
can  prove  that  —  =  -7-  =  — " 
a      b     c 


Ex.  4.     Show  that  Prob.  15  may        a 
be  solved  as  follows  (Fig.  294) :     From  c 

A  draw  ray  /  and  from  B  draw  m\\l. 
Lay  off  the  given  segments  on  /  and  7n 
as  shown  in  the  figure  and  join  the 
pomts  of  division.  Fig.  294 


J--- 


RATIO  AND  PROPORTION  173 

CONSTRUCTION   OF  FOURTH  PROPORTIONALS 
207.  Ex.  1.     Given  three  segments  a,  b,  and  c,   construct   a 

fourth  segment  x  so  that  -7  =  -  (Fig.  295).        « 

^       *  c 

Suggestion.    The  construction  is  based  on  o^*:::— 2 — ^  — ^ ^ 

Th.  99.     How  are  the  segments  a,  b,  and  c  laid 
off  on  the  sides  of  Z  0? 

Ex.  2.    Using  the  three  segments  given  pj^   295 

a     b 
in  Fig.  295,  construct  x  so  that  —  =  — ,   so 
^  '  c     X  ^ 

be  ,       c     b         ^i.  J.   b    a  *      c     a 

that  -  =  -,  so  that-  =  -,  so  that  -  =  -  so  that-r  =  ~- 
ax  ax  ex  ox 

Ex.  3.  Do  any  of  the  figures  called  for  in  Ex.  2  give  the  same 
value  for  a:?     Why? 

The  fourth  term  of  a  proportion  in  which  the  other  three 
terms  are  the  three  given  numbers  taken  in  order  is  called 
the  fourth  proportional   to  the  three   given  numbers;  for 

example,  if  t=-,  ^  is  the  fourth  proportional  to  a,  b,  and  c. 

Ex.  4.     Find  the  fourth  proportional  to 

a.  21,  5,  and  4  e.  6H,  8Hj  and  5        e.  a+l,  a,  and  a+4 

b.  5a,  3a,  and  2b  d.  a,  a+1,  and  Qa^       f.  Qj^,  26,  and  35 

Problem  16.  To  construct  a  fourth  proportional  to  three 
given  segments. 

Analysts: 

Let  a,  b,  and  c  represent  the  given  segments  and  x  the 
fourth  proportional. 

To  construct  a  foiuth  proportional  to  a,  6,  and  c,  construct 

X  so  that  I  =  - .     (See  Fig.  295.) 

Ex.  5.  ,  Using  three  given  segments  a,  b,  ,and  c,  find  a  fourth 
proportional  to  a,  c,  and  b;  to  b,  c,  and  a;  to  c,  a,  and  b.  Are  the 
words  "in  order"  essential  in  the  definition  of  the  fourth  propor- 
tional?   Why? 


174  PLANE  GEOMETRY 

Ex.  6.  Find  by  geometry  a  fourth  proportional  to  the  segments 
whose  lengths  are  given  below.  In  each  case  verify  by  measure- 
ment and  computation. 

a.  4.2  cm.,  2.5  cm.,  37  cm. 

b.  3.5  cm.,  4.9  cm.,  2.5  cm. 

c.  3      cm.,  3.5  cm.,  4.2  cm. 

Ex.  7.     If  a,  b,  and  c  represent  three  given  segments,  find  a 

fourth  segment  x  so  that  {a)  x=^-;  (b)  x  =  — -;  (c)  x  =  %-- 

a  a  b 


MISCELLANEOUS   EXERCISES   INVOLVING 
RATIOS   AND   PARALLELS 

CD     1 
208.  1.  In  Fig.  296,  ^  =  3*     If  DE\\AB   and 

DF\\CB,  prove   that  DE  =  y^   AB.    What   is   the 

ratio  of  DF  to  CJ5?  ^  'f~^ 

Fig.  296 

2.  Construct  between  two  sides  of  a  triangle  a  segment  that 
shall  be  parallel  to  the  third  side  and  equal  to  %  of  the  third  side. 

3.  If,  in  Fig.  297,  ED  \\  CB  and  CF  \\  EB, 

AD_AB    AD_AB        ^  AB _AF 
prove  ^^-^^  AB-Jf'  ^^^  DB-Jp- 

A 

A  nalysis: 

To  prove  the  two  ratios  equal,  prove  them  each  equal  to  a  third  ratio. 

4.  A  line  is  drawn  through  the  intersection  of  the  medians 
of  a  triangle  cutting  two  of  the  sides  of  the  triangle  and  parallel 
to  a  third.     In  what  ratio  are  these  sides  divided?    Why? 

c 

5.  In  Fig.  298,  O  is  any  point  within  A  ABC, 
A'B'\\AB  from  A',  an  arbitrary  point  in  AO, 
and  intersects  OB  at  B' .  B'C  \\  BC  from  B'  and 
intersects  OC  at  C.  A'  and  C  are  joined.  Prove 
that  ^'C'MC.  -      ^^^^^^       - 

6.  Would  Ex.  5  be  true  if  point  0  were  outside  A^^C?  Draw 
the  figure  and  give  the  proof. 


RATIO  AND  PROPORTION  175 

7.  In  Fig.  299,  CO  is  an  arbitrary  segment  from  C  to  AB  and 

X  is  a  point  on  CO.     If  ■:r^  is  equal  to  a  given  ratio  ^ 

(for  example,  ^^),  find  the  locus  of  point  X  as  O 
moves  along  the  line  AB. 


8.  In  Fig.  300,  O  is  the  mid-point  oi  AB,  AK 
=  liAB,  and  KH  \\  OC.  What  is  the  ratio  of  CH 
to  CA?    Answer  this  question  \i  AK=%  AB.  /^^  (f — ^ 

Fig.  300 

9.  If,  in  Fig.  301,  AB  =  BC,  AD  =  CE,  and  FB^ 
BG,  prove  that  AC\\DE  \\  FG. 

10.  Prove  by  Th.  99  that  if  a  line  is  parallel  to 
the  base  of  a  triangle  and  bisects  one  side,  it  bisects /^^V^—f^ 
the  other  also.  Fig.  301 

11.  Prove  the  converse  of  the  theorem  quoted  in  Ex.  10  by 
Th.  100.  Cor. 

12.  Prove  by  Th.  98  that  if  a  line  is  parallel  to  the  bases  of  a 
trapezoid  and  bisects  one  side,  it  bisects  the  other  side  also.  Can 
the  converse  of  this  be  proved  by  proportion? 

13.  Given  any  angle  and  P,  any  point  within  it.  Draw  a  line 
through  P  meeting  the  sides  of  the  angle  in  two  points  M,  N,  such 
that  MP  =  2PN. — College  Entrance  Examination  Board,  Plane 
Geometry  Examination,  1912. 

14.  Show  that  a  carpenter's  steel  square  may  be  used  to  solve 
problems  in  proportion.  Fig.  302  shows  a  steel  square  graduated 
to  half-inches.     ACD  is  a  frame   made   of 

two  pieces  of  wood  hinged  at  C.     AC  can 

slide  on  the  long  arm  of  the  square.     Find 

12    9 
a  number  x  so  that  —  =  —    Place  ACD  so 
16    X 

that  the  outer  edge  of  CD  is  on  6  on  the  long  p^^   3Q2 

arm  (12  half -inches)  and  4>^  on  the  short 

arm.    Without  changing  the  angle  of  adjustment  of  the  frame, 

move  the  frame  to  the  left  until  the  outer  edge  of  CD  passes 

through  8  on  the  long  arm.    How  is  the  value  of  x  found?    Why? 


176  PLANE  GEOMETRY 

SIMILAR  TRIANGLES 
TEST  I  FOR  SIMILAR  TRIANGLES 
209.  Theorem  101.     If  two  triangles  have  the  angles  of 
one  respectively  equal  to  the  angles  of  the  other,  the  corre- 
sponding sides  have  equal  ratios. 


Hypothesis:     AABC  and  ADEF  are  any  two  triangles 
with  ZA=ZD,  ZB=ZE,  and  ZC=  ZF. 

Conclusion :     — — = — —  =  — - . 
DE    EF    FD 

Analysis  and  construction: 

AT?         T^  C 

I.  To  prove  r—-=——,  use  a  line  parallel  to  a  third  side 

of  /\DEF. 
II.    .*.  place  /lABC  upon  ADEF  with  point  B  on  point  E, 
AB  along  DE,  and  BC  along  EF.     Then  prove 
A'a  II  DF. 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 

T^  •  1      ^  BC     CA 

It  IS  necessary  also  to  prove  — -  =  — —  • 

EF     FD 

Exercise.     If  a  and  a',  b  and  b',  c  and  c'  are  corresponding 
sides  of  mutually  equiangular  triangles,  find 

1.  ^j'andc',  if  o  =12,      6=18,    c  =24,  anda'  =  20. 

2.  yandc',  if  a  =3M,    6=4,      c  =5>i  and  a' =  25. 

3.  b  and  c,  it  a  =  lOH,  a'  =  16,    b'  =  20,  and  c'  =  27. 

210.  Two  polygons  are  said  to  be  similar  if 

a.  The   angles  of  one   are  equal   to   the  corresponding 

angles  of  the  other  and 

b.  Corresponding  sides  have  equal  ratios. 


RATIO  AND  PROPORTION  177 

Theorem  101  gives  us  our  first  test  for  similar  triangles. 
We  will  state  it  formally  as 

Theorem  102.  Two  mutually  equiangular  triangles  are 
similar.    Why? 

In  popular  language,  similar  figures  have  the  same 
shape.  All  enlargements  and  drawings  to  scale  are  practical 
examples  of  similar  figures. 

Ex.  1.     Prove  that  all  equilateral  triangles  are  similar. 

Ex.  2.  Are  all  isosceles  triangles  similar?  Why?  When  are 
two  isosceles  triangles  similar? 

Ex.  3.  If  two  isosceles  triangles  have  equal  vertex  angles,  the 
legs  have  the  same  ratio  as  the  bases. 

Ex.  4.  Are  all  right  triangles  similar?  Why?  When  are 
two  right  triangles  similar? 

Ex.  5.     Is  a  square  similar  to  a  rectangle?    Why? 

Ex.  6.     Construct  two  rectangles  that  are  similar. 

TESTS   FOR  EQUAL  PRODUCTS   AND   EQUAL  RATIOS 

211.  At  the  beginning  of  the  course  in  geometry  consid- 
erable time  was  spent  on  the  use  of  congruent  triangles  in 
proving  segments  and  angles  equal.  So  important  is  this 
that  when  it  is  necessary  to  prove  two  segments  equal  we 
often  look  first  for  a  pair  of  congruent  triangles.  In  this 
chapter  we  are  studying  especially  equal  ratios,  which  are 
just  as  important  as  equal  segments.  The  tests  for  equal 
ratios  are  as  important  as  the  tests  for  equal  segments. 
When  two  ratios  are  to  be  proved  equal,  the  following  possi- 
bilities must  be  considered: 

A.  Our  fundamental  methods  for  proving  ratios  equal  are: 

1.  By  parallels  and  transversals. 

2.  By  similar  triangles. 

B.  Before  either  of  these  mechods  can  be  applied  it  is 
often  necessary  to  find  a  third  ratio  to  which  each  of  thei 
given  ratios  can  be  proved  equal. 


178  PLANE  GEOMETRY 

The  use  of  similar  triangles  in  proving  ratios  equal  is  of 
considerable  importance.  The  following  considerations  are 
often  helpful : 

First:  To  select  the  proper  triangles:  The  definition  of 
similar  figures  says  that  corresponding  sides  have  equal 

ratios.    This  gives  —  =  77  and  by  alternation  ir  =  77  *     If. 

1  c       e 

then,  we  are  to  prove  two  ratios  equal,  say  -7-  =  y ,  we  may 

choose  the  triangles  so  that  one  of  them  shall  have  c  and  e 
as  sides  and  the  other  shall  have  d  and  /  as  sides,  or  so  that 
one  of  them  shall  have  c  and  d  as  sides  and  the  other  shall 
have  e  and  /  as  sides ;  that  is  so  that  the  numerators  shall 
be  sides  of  one  triangle  and  the  denominators  sides  of  the 
other,  or  so  that  the  terms  of  one  ratio  shall  be  sides  of  one 
triangle;  and  the  terms  of  the  other  ratio  sides  of  the  other. 

Second:  To  select  the  corresponding  sides  of  a  pair  of 
similar  triangles:  In  Fig.  303  the  triangles  are  so  placed 
that  corresponding  sides  can  be  selected  immediately  by 
inspection.  When  the  triangles  are  not  thus  conveniently 
placed  it  is  necessary  to  remember  that  corresponding  sides 
are  always  opposite  equal  angles.  The  corresponding  sides 
should  be  selected  carefully  from  the  equal  angles  as  illus- 
trated in  the  proof  to  Ex.  1  on  p.  179.  Notice  that  equal  angles 
may  be  designated  by  the  same  numbers,  as  Z2  and  Z2'. 

When  it  is  required  to  prove  two  products  equal,  a  pair 
of  equal  ratios  may  be  obtained  from  the  equal  products  by 
Th.  91  and  the  ratios  proved  equal  as  explained  above. 

Note.     If  we  say  that  corresponding  sides  of  similar  triangles  have 

a        b 
equal  ratios,  the  ratios  should  be  read  ~' '^  ~y'   If>  however,  we  say  that 

corresponding  sides  of  similar  triangles  are  proportional,  we  may  use 

either  -^  =  -77  or  -?-  =  ■^-     Similarly  in  Fig.  289  if  we  say  the  sides  are 
a       0        0       0 

divided  proportionally,  we  may  use  the  ratios  given  in  Th.  99  and  Cor. 
or  either  of  the  forms  in  Ex.  1,  §  204. 


RATIO  AND  PROPORTION  179 

EXERCISES  INVOLVING  THE  USE   OF  TEST  I  FOR 
SIMILAR  TRIANGLES 

212.  1.  The  diagonals  of  a  trapezoid  divide  each  other  into 
segments  that  have  the  same  ratio  (see  Fig.  304). 

Analysis: 


X     y 

To  prove  — =  —  prove  the  angles  of  ADOC  equal  y 

respectively  to  the  angles  of  AOBA.  ^^^'  ^O^ 

Proof: 

STATEMENTS 

1.  Z1=Z1'. 

2.  Z2=Z2'. 

3.  ADOC=ZAOB. 

.    X  (opposite  Z2)  _  y  (opposite  Zl)  ^ 
z  (opposite  Z2')     zy  (opposite  Zl') 

2.  In  Fig.  305,  0  is  any  point  in  segment  AB.     y^ 

Any  line  is  drawn  through  point  O  not  perpen-     \^ 
dicular  to  AB.     From  A  and  B  perpendiculars        j^ 

are  drawn  meeting  this  line  at  points  F  and  X.         Fig,  305 

3.  A  BCD  is  a  parallelogram  with  its  diagonal  AC.  BX  is  a 
line  drawn  through  B  intersecting  AC  at  Y  and  ^D  at  X.  Prove 
;,    ^  BY    BC 

'^"'  XY  =  A-X 

4.  In  Fig.  306,  ABC  is  an  isosceles  triangle.        y^^^^^ 

Z  1  =  Z  2.     Prove  that  b^  =  cm.  ^^'.k..J^.B 

Analysis:  F^c.  306 

I.  To  prove  b^  =  cm,  prove  that  -r=  — 

0     tn 

If  the  product  of  two  segments  equals  the  square  of  a 
third  segment,  the  last  segment  is  called  a  mean  propor- 
tional between  the  other  two.     In  Ex.  4,  h'^  =  cm,  6  is  a  mean 

proportional  between  c  and  m.     If  —  =-r'   —=—iX^  =  ab, 

OC         0        d        % 

or  X  =  Va6,  :!c  is  a  mean  proportional  between  a  and  h.    Why  ? 

5.  Investigate  the  case,  Fx.  4,  in  which  A  A  >  Z.C. 


180 


PLANE  GEOMETRY 


6.  In  Fig.  307,  AABC  is  a  right  triangle  with   Z5  a  rt.  Z. 
DE  is  drawn  ±  AC  from  any  point  in  AB.     Prove 
AB  'AD  =  AE  'AC. 

7.  Investigate  the  case,  Ex.  6,  in  which  point  D 
is  on  AB  extended.  (. 

Fig.  307 

8.  Investigate  the  case,  Ex.  6,  in  which  point  D  is  on  BA 
extended. 


9.  Prove  that  the  parallel  sides  of  a  trapezoid  have  the  same 

ratio  as  the  segments  into  which  one  diagonal  is  divided  by  the 

other. 

c 

10.  In  Fig.  308,  AABC  is  isosceles  and  BX  =  BA.        ^> 
Prove  that  c  is  a  mean  proportional  between  AC 
and  AX, 


11.  In  Fig.   d09,CX±AB  and  BY±AC.     Prove 
^,       AC     CX 

^^^^AB  =  BY' 

12,  In  Fig.  309,  prove  that  BO  ■  BY=^BA  •  BX. 


13.  In  Fig.  310,  lines  h  and  k  are  parallel  and  are 
cut  by  the  pencil  of  rays  from  point  0.     Prove  that 
a  _c 
~b~l' 

Suggestion.     Prove  that  each  ratio  is  equal  to  a  third 
ratio. 


A  X    B 

Fig.  309 


Fig.  310 


14.  What  ratios  would  be  equal  if  h  and  k  (Fig.  310)  were 
on  opposite  sides  of  point  O  ? 


15.  In  Fig.  311,  AABC  is  a  right  triangle 
with  Z  C  a  right  angle.    CD±AB  from  C.    Prove      y 
AACD^ACBD  and  read  the  ratios  of  corre- ^. 
spending  sides. 


Fig.  311 


RATIO  AND  PROPORTION 


181 


16.  In  Fig.  312,  ^B  is  a  diameter  of  OO,  BD  tangent 
circle  a.t  B.     AD  is  any  line  from  A  cutting  the  circle 
at  E  and  the  tangent  at  D.     Prove  that  AB  \s2i  mean 
proportional  between  AE  and  AD. 


17.  In  Fig.  312  draw  BE.  Prove  that  BE  is  a 
mean  proportional  between  AE  and  ED. 

18.  In  Fig.  313,  CZ)  is  a  diameter  perpendicular  to 
chord  AB.  Prove  that  ^C  is  a  mean  proportional 
between  CE  and  CD. 

19.  In  Fig. '313  prove  that  /IE  is  a  mean  propor- 
tional between  CE  and  ED. 

20.  In  Fig.  313  prove  that  CE-ED  =  AE^EB. 

21.  The  cross-section  of  a  street  surface  is  the  arc 
of  a  circle;  the  distance  from  curb  to  curb  is  30  ft.; 
the  rise  of  the  center  of  the  street  above  the  gutter  is 
7  inches.    What  is  the  radius  of  the  circle? 

22.  In  Fig.  314,  XF  II  ^5  and  FZ II  BC.  Prove  that 
b~l' 


Fig.  313 


Fig.  314 


23.  Draw  a  square  A  BCD  and  the  diagonals  AC  and  BD. 
Let  E,  F,  G,  and  H  be  the  mid-points  of  the  sides  AB,  BC,  CD, 
and  DA  respectively.  Join  each  vertex  to  the  mid-points  of  the  two 
non-adjacent  sides,  that  is,  join  A  to  points  F  and  G,  and  so  on. 
Find  pairs  of  similar  triangles  and  read  the  ratios  of  corresponding 
sides. 


24.  Fig.    315    shows    three    concurrent    lines 


XY\\AB,  and  YZ\\BC.    Prove  that  j  =  y  • 

25.  In  Fig.  316  the  circles  are  tangent  at  X, 
AB  and  CD  are  drawn  through  the  point  of  tan- 
gency,  meeting  the  circles  as  shown.  Prove  that 
cb  =  ad.     (See  §175,  Ex.20.) 

26.  Investigate  the  case,  Ex.  25,  in  which  the 
circles  are  tangent  internally. 


Fig.  316 


13 


182  PLANE  GEOMETRY 

IMPORTANT   SPECIAL   CASES 
INTERSECTING   CHORDS 

213.  Theorem  103.  If  two  chords  intersect  within  a 
circle,  the  product  of  the  segments  of  one  is  equal  to  the 
product  of  the  segments  of  the  other. 


Fig.  317 

Hypothesis:  Circle  O  is  any  circle  with  the  chords  AB 
and  CD  intersecting  at  X  so  that  a  and  h  are  the  segments 
of  AB  and  c  and  d  the  segments  of  CD. 

Conclusion :     ab  =  cd. 

The  analysis,  construction,  and  proof  are  left  to  the  pupil. 

In  the  next  six  exercises  the  letters  refer  to  Fig.  317. 

Ex.  1.  Find  b,  if  a  =  12,  ^  =  2%,  and  c=15. 

Ex.  2.  Find  a  and  b,iiAB  =  22,d  =  8,  and  c  =  12. 

Ex.  3.  Find  a  and  d,  if  AB  =  19,  6  =  10,  and  c  =  6. 

Ex.  4.  Find  d,  if  a  =  5>^,  <;  =  43^,  and  b  =  5. 

Ex.5.  I^mda,iid  =  8li,c  =  2%,sindb  =  3y5, 

Ex.  6.  Find  c  and  d,iiAB  =  2Q,b  =  8,  and  c  =  d. 

214.  If  point  C  is  between  A  and  B  on  line  AB,  AC  and 
C-5  are  said  to  be  segments  oiAB  and  ^B  is  said  to  be  divided 
internally  at  C.     In  Fig.  318,  AB  is  divided  ^         c   ^ 
internally  at  C.     ylC+C5  =  A5.  i -5" — c' 

If  point  C  is  on  line  AB  but  not  between 
A  and  5,  AC  and  -BC  are  still  said  to  be  segments  of  AB. 
AB  is  said  to  be  divided  externally  at  C.     In  Fig.  318, 
AB  is  divided  externally  at  C.     AC'-BC'  =  AB  if  C  is 
on  AB  extended.     BC'-AC'=BA  if  C  is  on  BA  extended. 


RATIO  AND   PROPORTION 


183 


INTERSECTING   SECANTS 

215.  Theorem  104.  If  two  secants  intersect  without  a 
circle,  the  product  of  one  secant  and  its  external  segment 
is  equal  to  the  product  of  the  other  secant  and  its  exter- 
nal segment. 


Fig.  319 


Hypothesis:  Circle  O  is  any  circle  with  the  two  secants 
h  and  k  intersecting  without  the  circle  at  E  so  that  a  and  b 
are  the  external  segments  of  h  and  k  respectively. 

Conclusion:    ah  =  bk. 

The  analysis,  construction,  and  proof  are  left  to  the  pupil. 

Ex.  1.     The  following  data  refer  to  Fig.  319;  c  and  d  are  the ' 
internal  segments  of  h  and  k  respectively.     Find  the  length  of  the 
segments  required. 

a.  Find  c  and  d,  if  A=15,  a  =  7,  and  ^  =  35. 

b.  Find  b,  if  a  =  9,  k=  12,  and  c  =  4. 

c.  Find  a,  if  <f  =  8,  c=  13,  and  6  =  4. 

d.  Find  a,  if  c=10,  b  =  4,  and  d  =  20. 

Ex.  2.  If  in  QO  the  chords  AB  and  CD  intersect  at  X  (Fig. 
320),  what  kind  of  segments  are  AX,  BX, 
CX,  and  DX?  If  point  B  moves  along  the 
circle  until  AB'  intersects  CD  without  the 
circle  at  X',  what  kind  of  segments  are  AX', 
B'X',  CX',  and  DX'?  Show  that  the  chords 
are  divided  in  one  case  internally  and  in  the 
other  case  externally  so  that  the  product  of  the  segments  of  one 
chord  is  equal  to  the  product  of  the  segments  of  the  other. 


Fig.  320 


184  PLANE  GEOMETRY 

INTERSECTING  TANGENT  AND   SECANT 

216.  Theorem  105.  If  a  secant  and  a  tangent  meet 
without  a  circle,  the  tangent  is  a  mean  proportional  between 
the  whole  secant  and  its  external  segment. 


Fig.  321 

Hypothesis:  Circle  0  is  any  circle  with  secant  k  and 
tangent  h  meeting  without  the  circle  at  C  so  that  a  is  the 
external  segment  oi  k. 

Conclusion:    h'^=ak. 

The  analysis,  construction,  and  proof  are  left  to  the  pupil. 

In  the  next  five  exercises  the  letters  refer  to  Fig.  321. 

Ex.  1.     Find  k,  ii  a  =  4:  a.nd  h  =  6. 

Ex.  2.    Find  kanda^ii  AB  =  7  and  h  =  12. 

Ex.  3.  Find  a,  if  h=10,  the  radius  of  the  circle  is  7j^,  and  the 
secant  passes  through  the  center  of  the^circle. 

Ex.  4.     Find  k  and  a,  if  .45  =  27  and  h  =  18. 

Ex.  5.  Find  the  radius  of  the  circle  if  a  =  4,  h  =  l2,  and  the 
secant  passes  through  the  center  of  the  circle. 

Ex.  6.  In  Fig.  320,  what  motion  of  AX'  will  show  the  relation 
between  Ths.  104  and  105? 

Ex.  7.  Draw  a  segment  AC=S  cm.  On  AC  from  C  lay  off 
CB  =  2  cm.  Draw  any  circle  passing  through  points  A  and  B. 
Draw  a  tangent  to  this  circle  from  point  C.  Draw  several  figures, 
varying  the  radius  of  the  circle  drawn,  but  using  always  ^C  =  S  cm. 
and  C5  =  2  cm.  Compare  the  lengths  of  the  various  tangents. 
Can  you  explain  the  results? 

Ex.  8.  Tangents  drawn  to  two  intersecting  circles  from  any 
point  in  their  common  chord  are  equal. 


RATIO  AND   PROPORTION  185 

SEGMENTS   MADE  BY  THE  BISECTOR   OF  AN   ANGLE 
OF   A   TRIANGLE 

217.  Theorem  106.  The  bisector  of  an  angle  of  a  tri- 
angle divides  the  opposite  side  internally  into  segments 
that  have  the  same  ratio  as  the  other  two  sides  of  the 
triangle. 


Fig.  322 

Hypothesis:     AABC  is  any  triangle  with  CO  bisecting 
ZBCA  and  dividing  AB  into  segments  r  and  s. 

Conclusion:  —  =  t  • 
5      b 

Analysis  and  construction: 

I.  Two  ratios  may  be  proved  equal  by  use  of  Th.  98, 

Th.  99,  or  Th.  101.     (Why?)     We  will  use  Th.  99. 

T      a 
II.    .*.  construct  AD  from  A  \\  OC  and  get  —  =  r,' 

Y      a  .  T      a 

III.  To  prove  —  =  t,  prove  h  =  h'  and  substitute  in  -  =  j-,- 

'Let  the  pupil  complete  the  analysis  and  give  the  proof. 
In  the  next  six  exercises  the  letters  refor  to  Fig.  322. 


Ex.  1. 

a  =  12, 

b=  9, 

r=  5, 

find  s. 

Ex.2. 

a  =  72, 

6  =  60, 

r  =  32, 

find  s. 

Ex.3. 

a  =  28, 

6  =  21, 

AB  =  Zb, 

find  r  and  s. 

Ex.4. 

r=15, 

5=18, 

a  =  40, 

find  6. 

Ex.5. 

r=  8, 

s=   9, 

6  =  34, 

find  a. 

Ex.6. 

a  =  18, 

6  =  16, 

^5  =  24, 

find  r  and  s. 

Ex.7. 

Discuss 

the  case,  Th.  106, 

in  which  A> 

Prove  by  proportion. 


186  PLANE  GEOMETRY 

SEGMENTS   MADE  BY   THE  BISECTOR   OF  AN 
EXTERIOR  ANGLE   OF  A  TRIANGLE 

218.  Theorem  107.  The  bisector  of  an  exterior  angle  of 
a  triangle  divides  the  opposite  side  externally  into  segments 
that  have  the  same  ratio  as  the  other  two  sides  of  the  triangle. 


Fig.  323 

Hypothesis:  AABC  is  any  triangle  with  CO  bisecting  the 
exterior  ZACE  and  dividing  BA  externally  into  segments 
r  and  5.     Let  BC  =  a,  CA  =  b,  and  CD  =  b'. 

Conclusion :      L-^. 
s     h 

The  analysis,  construction,  and  proof  are  left  to  the  pupil. 

Discussion.  Theorem  107  is  not  true  for  the  vertex  angle 
of  an  isosceles  triangle. 

In  the  next  four  exercises  the  letters  refer  to  Fig.  323. 

Ex.  1.     Find  a,  if  r  =  24,  5  =  6,  and  6  =  5. 

Ex.  2.    Find  r,iia=  18,  6  =  5,  and  ^  =  7. 

Ex.  3.     Find  r  and  s,  if  AB=S,  a=12,  and  &  =  6. 

Ex.  4.     Find  r  and  s,  if  AB  =  4:,  a  =  9,  and  b  =  6. 

219.  If  two  points  divide  a  segment  internally  and  exter- 
nally in  the  same  ratio,  the  segment  is  said  to  be  divided 
harmonically  by  the  two  points.  ^^^ 

Ex.  1.     In  Fig.  324,  ABC  is  any  triangle.  ^ 

CX  bisects  ZACB  and  CX'  bisects  the  exterior 
/.BCD.     Prove  that  AB  is  divided  harmoni-  ^      ^  ^ 
cally  at  X  and  X'.  ^'^-  ^^4 

Ex.  2.     Show  how  to  divide  any  given  segment  harmonically. 

Suggestion.     Construct  any  triangle  on  the  segment  AB  as  base. 


RATIO  AND   PROPORTION 


187 


PROPORTIONAL   SEGMENTS   IN   RIGHT   TRIANGLES 

220.  Theorem  108.  If  a  perpendicular  is  drawn  from  the 
vertex  of  the  right  angle  of  a  right  triangle  to  the  hypotenuse, 
the  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse. 


Fig.  325 

Hypothesis:     A  ABC   is    any   triangle    with    ZC  =  rt.  Z, 
p±  AB  from  C.     AB  =  c\  m  and  n  are  the  segments  of  c. 
Conclusion:     ^  is  a  mean  proportional  between  m  and  n. 
The  analysis  and  the  proof  are  left  to  the  pupil. 

Theorem  109.  If  a  perpendicular  is  drawn  from  the  ver- 
tex of  the  right  angle  of  right  triangle  to  the  h3rpotenuse, 
either  leg  is  a  mean  proportional  between  the  whole  hypote- 
nuse and  the  segment  adjacent  to  that  leg. 

The  analysis  and  the  proof  are  left  to  the  pupil. 

Exercise.  Construct  Fig.  325  so  that  (a)  m  =  Z  cm.  and  «  =  7 
cm.;  (6)  6  =  6  cm.  and  c  =  9  cm.;  (c)  6  =  2.4 cm.  and  w  =  1.2  cm.  In 
each  case  measure  the  remaining  segments  and  compare  them  with 
the  results  obtained  by  computation. 

CONSTRUCTION   OF  MEAN  PROPORTIONALS 

221.  Problem  17.  To  construct  a  mean  proportional 
between  two  given  segments. 

Solution  I.  The  solution  may  be  based  on  Fig.  326,  by  making  the 
hypotenuse  of  the  right  triangle  equal  to  the  sum  of  the  two  given 
segments.     Construct  the  right  triangle  by  means  of  a  semicircle. 

Solution  II.     The   solution  may  be  based  on  ^ 

Fig.  326,  by  making  c  equal  to  the  longer  and  m 
equal  to  the  shorter  of  the  given  segments. 


Can  you  invent  a  solution  based  on  Fig.  321? 


Fig.  326 


188  PLANE  GEOMETRY 

Ex.  1.     If  a  and_b  are  two  given  segments,  construct  a  segment 
X  so  that  (1)  x=  ylah;  (2)  x=  ^2ab;  {S)x=  V^^a^;  (4)  x  =  }^  ^|ab, 

Ex.  2.   JTaking  any  given  length  to  represent  1,  find  segments 

equal  to  V2,  V8,  V12.     Measure  the  results  and  compare  them 

with  the  approximate  square  roots  of  2,  8,  and  12. 

—  2       X 

Suggestion.     If  x=yl2,x'^  =  2.    Then  "  =  ^* 

RELATION  BETWEEN  THE  SIDES  OF  A  RIGHT  TRIANGLE 

222.  Theorem  110.     The  sum  of  the  squares  of  the  legs 
of  a  right  triangle  is  equal  to  the  square  of  the  hypotenuse. 


Fig.  327 

Hypothesis:  AABC  is  any  right  triangle  with  ZC  a 
rt.  Z ,  c  the  hypotenuse,  and  a  and  b  the  legs. 

Conclusion:    a^-\-b^=c'^. 

Analysis  and  construction: 

I.  To  prove  a^-\-¥  =  c^,  find  a  value  for  a^  and  a  value 
for  b^  and  add. 

II.  The  terms  a^  and  b^  above  suggest  the  use  of  the  mean 

proportional  theorem. 
III.    .*.  draw  a  perpendicular  from  C  to  AB  and  find  a^ 

and  b^  in  terms  oi  AB  and  its  segments. 
Proof: 

STATEMENTS 

1.  a^  =  nc  and  b^  =  mc. 

2.  a^-]-b^  =  nc-]-mc={n-\-'m)c. 

3.  a2 +62  =  ^2. 

Let  the  pupil  give  the  reasons.  For  2  use:  the  sum  of  numbers 
having  a  common  factor  is  the  common  factor  multiplied  by  the  sum 
of  the  coefficients. 


RATIO  AND  PROPORTION  189 

Note.  Th.  110  is  one  of  the  most  important  theorems  of  geometry 
and  one  of  tlie  most  frequently  used.  More  than  one  hundred  proofs 
are  known.  It  is  called  the  Pythagorean  theorem.  Pythagoras,  a 
Greek,  is  supposed  to  have  given  a  general  proof,  although  the  fact  was 
believed  to  be  true  much  earlier.  We  do  not  know  the  nature  of  the 
proof  that  Pythagoras  gave,  but  it  is  probable  that  it  was  something 
like  the  one  given  above.  Pythagoras  settled  in  Crotona,  Southern 
Italy,  where  he  founded  a  brotherhood,  the  members  of  which  were 
pledged  to  secrecy.  They  spent  their  time  in  the  study  of  philosophy, 
ethics,  and  mathematics  and  discovered  many  important  theorems 
in  geometry.     Pythagoras  died  about  501  B.C. 

Exercise.  What  is  the  hypotenuse  of  a  right  triangle  if  the 
perpendicular  sides  are  3  and  4? 

Note.  The  fact  that  a  triangle  whose  sides  are  3,  4,  and  5  is  a 
right  triangle  and  the  relation  3^+42  =  52  were  known  to  the  Egyptians 
more  than  three  thousand  years  before  the  time  of  Pythagoras. 
The  Egyptians  used  ropes  knotted  at  equal  distances.  These  were 
stretched  about  three  poles  so  as  to  form  a.  right  triangle.  The 
Egyptians  called  the  men  who  knew  how  to  use  these  ropes  rope- 
stretchers.  Surveyors  use  similar  methods  to-day.  How  many  knots 
must  there  be  in  the  rope  and  how  is  it  used  to  construct  a  right 
triangle? 

The  pyramids  of  Egypt  have  an  angle  nearly  equal  to  an  acute 
angle  of  a  triangle  whose  sides  are  3,  4,  and  5.  The  Chinese  and 
the  Hindus  probably  knew  about  this  right  triangle  at  a  very  early 
date. 

APPLICATIONS   OF  PYTHAGOREAN   THEOREM 

223.  Ex.  1.  What  is  the  diagonal  of  a  rectangle  whose  sides 
are  25  ft.  and  60  ft.? 

Ex.  2.  The  diagonal  of  a  rectangle  is  50  ft.;  one  side  is  14  ft. 
Find  the  other  side. 

Ex.  3.    Find  all  of  the  segments  in  Fig.  328,  if  ^ 

a.  «  =  16,     p=12.       d.   6=10,    a  =  24. 

b.  m=  8,     6=17.        e.    c=50,     6  =  30. 

c.  p=  9,    w=12.        /.  w  =  28,    »  =  63. 
Ex.  4.     The    radius    of    a    circle    is    11    in. 

Find  the  length  of  a  tangent  drawn  from  a  point  61  in.  from  the 
center. 


190  PLANE  GEOMETRY 

Ex.  5.  In  a  given  circle  let  r  represent  the  radius,  c  a  chord, 
and  d  the  distance  of  the  chord  from  the  center.  Find  the  missing 
terms  as  indicated  below: 

a,  d  =  8,        c  =  32,       r=?  c.  d=12,     c=?,     r  =  36. 

b.  d=?,        c=28,      r=18.       d.  d  =  5,      c=?,    r=45. 

Ex.  6.  One  side  of  a  square  is  12.  Find  the  sides  of  an  isosceles 
triangle  formed  by  joining  the  mid-point  of  one  side  to  the  opposite 
vertices. 

Ex.  7.  One  side  of  a  square  is  14.  Find  the  sides  of  an  isosceles 
triangle  formed  by  joining  one  vertex  to  the  mid-points  of  the  sides 
not  passing  through  that  yertex. 

Ex.  8.  If  one  side  of  a  square  is  8,  find  the  diagonal.  Find  the 
diagonal  if  one  side  is  6;  10;  12;  20;  5. 

Ex.  9.     If  a  diagonal  of  a  square  is  15,  find  the  side. 

Ex.  10.  If  the  hypotenuse  of  an  isosceles  right  triangle  is  12, 
find  the  side. 

Theorem  HI.  If  one  side  of  a  square  is  s,  its  diagonal 
is  s  V2.    If  the  diagonal  of  a  square  is  d,  the  side  is  ^/id  yl2. 

Ex.  11.  The  base  of  an  isosceles  triangle  is  12;  the  equal  sides 
are  18.    What  is  its  altitude? 

Ex.  12.  Find  one  of  the  legs  of  an  isosceles  triangle  if  the  alti- 
tude is  8  and  the  base  is  12. 

Ex.  13.  Find  the  altitude  of  an  equilateral  triangle  one  of 
whose  sides  is  4.     Find  the  altitude  if  the  side  is  5;  10;  12;  5. 

Ex.  14.  If  the  altitude  of  an  equilateral  triangle  is  14,  find  the 
side.     Find  the  side  if  the  altitude  is  6;  8;  7>^;  a. 

Theorem  112.  If  one  side  of  an  equilateral  triangle  is  s, 
its  altitude  is  V2  s  ^3.  If  the  altitude  is  a,  one  side  of  the 
equilateral  triangle  is  ^/s  a  VJ. 

Ex.  15.  One  side  of  a  rhombus  is  24;  one  of  its  angles  is  60°. 
Find  the  diagonals. 

Ex.  16.  One  angle  of  a  rhombus  is  60°.  The  longer  diagonal 
is  12.     Find  the  side. 


RATIO  AND  PROPORTION  191 

APPLICATIONS   OF   EQUAL   RATIOS 
MISCELLANEOUS  EXERCISES 

224.  1.  Fig.  329  shows  proportional  compasses  used 
to  enlarge  or  reduce  drawings  to  scale.  How  must 
the  instrument  be  adjusted  so  that  b  is  twice  a  ?  so  that 
b  is  three  times  a  ? 

2.  The  shadow  of  a  tree  is  36  ft.  at  the  same  time 

that  the  shadow  of  an  8-ft.  pole  is  5  ft.     How  high 

is  the  tree  ? 

Fig.  329 

3.  How  may  the  height  of  a  flagpole  be  found  by  noticing  just 
when  the  length  of  the  shadow  of  a  certain  post  is  equal  to  the 
height  of  the  post?    Why? 

Note.  It  is  said  that  the  Greek  Thales  astonished  the  Egyptians 
by  measuring  the  heights  of  the  pyramids  from  their  shadows.  Whether 
he  used  the  method  of  Ex.  2  or  the  special  case  mentioned  in  Ex.  3 
is  not  known.  Thales  lived  from  about  640  to  548  B.C.  and  introduced 
the  study  of  geometry  into  Greece.  x 

4.  An  8-ft.  pole  is  placed  at  B  (Fig.  330).     How  '' '^ 
may  point  A  be  located  and  what  lines  must  be  meas-       y 
ured  in  order  to  find  the  height  of  the  tower  CX?  Xb  ^ 
Why?                                                                                                 Fig.  330 

5.  Can  you  use  the  method  of  Ex.  4  in  determining  the  length 
of  a  flagpole  placed  on  the  comer  of  a  building?  Could  the  length 
be  determined  by  using  the  special  method  of  Ex.  3? 

6.  Show    that  an  inaccessible    distance    AB  (Fig.  331)   may 
be  obtained    by  the    following    method:     Make  c 
AC  LAB.    Take  C,  any  point  on  .4  C  from  which 
B  is  visible.     Make  CD  ±  CB.    Find  E,  the  point    ^' 
at  which   CD  and   BA   would  intersect.    What 
lines  must  be  measured? 

7.  Show  how  the  measurement  suggested  in  Ex.  6  may  be 
performed  practically  by  the  aid  of  a  pole  and  a  carpenter's  steel 
square. 

Note.  In  Ex.  6  stakes  are  set  up  at  the  points  indicated  and  the 
figure  laid  out  on  the  ground.  In  Ex.  7  the  figure  is  set  up  in  a  vertical 
plane. 


192 


PLANE   GEOMETRY 


8.  Show  that  an  inaccessible  distance  AB  (Fig.  332)  may  be 
obtained  as  follows:  Make  CA  ±  AB  and  extend 
AC  until  CD  is  some  convenient  part  oi  AC  i}4 
or  H).  Make  DE±AD.  Find  E,  the  point  at 
which  BC  and  DE  would  intersect.  What  measure- 
ment must  be  taken?  Is  it  necessary  that  AB  and 
DEhe±AD?    Why? 


Fig.  332 


9.  Before  the  invention  of  the  telescope  an  instrument  called 
the  cross-staff  was  sometimes  used  to  measure  inaccessible  heights 

and  distances.    The  cross-bar  c  was  made  to  slide  Tp^--^ 

up  and  down  the  staff  a  (Fig.  333).     Show  how  this  -^ ^       ''^ 

instrument  could  be  used  to  find  the  width  of  the       Fig,  333 
stream  R.    How  could  it  be  used  to  find  the  height  of  a  steeple? 


V  TRIGONOMETRIC  RATIOS 

225.  From  §210  we  know  that  if 


AABCc^AA'B'C  (Fig.  334)  ^  =  y 
or  T  =  T}'  Why?     The  second  pro-  ^ 


b     h' 

portion  may  be  translated  thus:  The  ratio  of  two  sides  of 
one  of  two  similar  triangles  equals  the  ratio  of  the  two  corre- 
sponding sides  of  the  other.     In  right  triangles  we  have: 

If  two  right  triangles  have  an  acute  angle  of  one  equal  to 
an  acute  angle  of  the  other,  the  ratio  of  any  two  sides  of 
one  is  equal  to  the  ratio  of  the  two  corresponding  sides  of 
the  other. 


Ex.  1.     In  Fig.  335,  AABC  and  A'B'C  are 

similar    right    triangles.     Apply    the    proportion 
given  above.     Obtain  three  pairs  of  equal  ratios. 

Note.  In  Fig.  335  the  triangles  are  lettered  so 
that  side  a  is  opposite  A  A,  side  b  is  opposite  Z.B, 
side  c  is  opposite  Z  C.  Angle  C  is  the  right  angle. 
For  convenience  in  the  discussion  that  follows,  the 
right  triangles  will  be  lettered  in  this  way. 


RATIO  AND  PROPORTION  193 

The  ratio  —  is  the  same  for  all  right  triangles  having  the 
c 

same  acute  angle  A.    Why?    The  same  may  be  said  of  the 
ratios  —  and  -r  •    We  may  say,  therefore,  that 

I.  If  an  acute  angle  of  a  right  triangle  is  known,  we  can 
find  the  ratios  of  the  sides. 

XL  If  the  ratio  of  any  pair  of  sides  of  a  right  triangle  is 
known,  we  can  find  the  angle. 

These  same  ratios  are  of  so  much  importance  in  relation 
to  the  /.A  that  they  have  been  given  special  names.  The 
names  of  the  ratios  are  as  follows: 

a        side  opposite  Z.A  .       „    ,    .        c    ,  a 

—  or  — :; — ^-^-: IS  called  sine  oi  Z.A   and  is  written 

c  hypotenuse 

sin  A. 

h        side  adjacent  to   ZA   .        ,,    -         .         r     ,  a 

—  or  T-^ — IS  called  cosine  of   /.A  and  is 

c  hypotenuse 

written  cos  A . 

a  side  opposite  ZA      .         „    .  ^  4.    r    y  a        j  • 

-r  or  -T-; TT-^ — 7— TH   IS  called  tangent  01   Z  A  and  is 

0        side  adjacent  to  ZA 

written  tan  A . 

Ex.  2.     Construct  two  right  triangles  with  angle  A  =  50°  in  each 

but  with  sides  of  different  lengths.     In  each  triangle  measure  a,  ft, 

I  -I  ,  .      a      b         .    a 

and  c  and  compute  the  ratios  — ,   — ,  and  -r- 

Ex.  3.  Follow  the  directions  given  in  Ex.  2  for  triangles  with 
angles  of  40°,  35°,  62°. 

Compare  the  results  obtained  in  Exs.  3  and  4  with  the  tables 
given  on  page  299  showing  the  values  of  these  ratios  for  angles 
of  all  degrees  from  1°  to  90°.  The  ratios  in  the  tables  are  given 
approximately  with  three  figures.  With  your  crude  methods  of 
measuring  you  cannot  expect  to  be  as  accurate. 

Q 

Ex.  4.  Construct  a  right  triangle  so  that  the  ratio  -r  is  %. 

Measure  ZA  with  a  protractor  and  compare  the  result  with  the 
table.     Look  for  the  tangent  that  is  nearest  H  and  note  the  angle. 


194  PLANE  GEOMETRY 

226.  It  is  important  to  notice  that  every  acute  angle  has 
a  particular  value  for  each  of  the  three  ratios.  If  the  angle 
is  given,  the  ratios,  sine,  cosine,  and  tangent  can  be  read 
from  the  table;  for  example, 

sin  23°  is  .391. 

Ex.  1.    Find  from  the  table 

sin  36°;  cos  42°;  tan  27°. 
sin  49°;  cos  15°;  tan  76°. 

So  also  if  the  ratio  is  given,  the  angle  can  be  found  from 
the  table. 

Ex.2.    What  is  Z^  if 

sin  ^  =  .515 ;  sin  ^  =  .966 ;  sin  ^  =  .777. 
cos  ^  =  .961 ;  cos  yl  =  .839  •,cosA  =  .292. 
tan  A  =  .325;  tan  ^  =  1.00;  tan  ^  =3.73. 

If  the  given  ratio  is  not  found  in  the  table,  use  the  one 
nearest  to  it;  for  example, 

if  sin  yl  =  .  239,  Z  A  is  about  14°, 
if  tan  A  =  1 .  56,  Z  A  is  about  57°. 

Ex.3.     Find  Z^  when 

sin  A  =  .472 ;  cos  ^  =  .395 ;  tan  ^  =  .726. 

227.  The  trigonometric  ratios  are  used  to  find  the  sides 
and  angles  of  right  triangles. 

Ex.  1.     In  A  ABC  (Fig.  336)  it  is  known  that 
a  =  23  and  (;  =  30;  find  ZA. 

Solution: 

Select  the  ratio  involving  the  opposite  side  and  the     / 

hypotenuse. 

Fig.  336 

sin  A  = — 
c 

_23 

"30 

=  .767  (by  division). 

.-.  Z.4  =50"  about  (from  table). 


RATIO  AND  PROPORTION  195 

Ex.2.  Find  side  b  of  the  AABC  when 
/.A  =35°  and  c=42  (Fig.  337). 

Solution: 

Select  the  ratio  involving  the  adjacent  side  and 
hypotenuse.  y^ 

cosA=—-  Fig.  337 

c 

cos  A  =cos  35°  =  .819  (from  table). 
.•.A=.819. 

Let  the  pupil  complete  the  solution. 

Ex.  3.  Find  side  b  when  a  =  42  and 
Z^=65°. 

Solution: 

Select  the  ratio  involving  the  opposite  side  and 
the  adjacent  side  (Fig.  338). 

tan  A  =-r ' 

0 

tan  A  =tan  65°  =  2.14  (from  table). 
.•.2.14=^-  Why? 

Let  the  pupil  complete  the  solution. 

These  exercises  illustrate  the  general  method  which  may- 
be stated  in  words:  To  find  any  particular  part  of  a  right 
triangle,  select  the  ratio  formula  involving  that  part  and  the 
two  known  parts;  form  an  equation  and  solve  it  for  the 
unknown  part. 

Ex.  4.     If  Zyl=64°  and  6  =  31,  find  c. 

Ex.  5.     li  ZA=  32°  and  a  =  54,  find  b. 

Ex.6.     If  Zyl  =  19°andc=16,  finda. 

Ex.  7.     If  a  =  35  and  c  =  47,  find  Z^l. 

Ex.  8.     If  6  =  52  and  c  =  73,  find  ZA. 

Ex.  9.     If  a  =  62  and  6  =  26,  find  ZA. 

228.  The  methods  illustrated  above  are  especially  used 
in  solving  problems  involving  heights  and  distances.  A 
tape  for  measuring  distances  and  some  instrument  for 
measuring  angles  are  needed  to  get  the  necessary  data 


T 


196  PLANE   GEOMETRY 

For  measuring  angles  the  surveyor  uses  a  transit  which  is 
really  two  protractors,  a  leveling  tube,  and  a  telescope  for 
easy  and  accurate  seeing.  For  accurate  work  he  measures 
angles  to  the  nearest  minute  or  even  closer  and  uses  more 
extended  ratio  tables.  Rough  approximations  may  be 
made  with  instruments  that  any  ingenious  pupil  can  make. 
A  good  protractor,  a  plumb  line,  and  a  couple  of  pegs  for 
sights  are  all  that  is  needed  for  a  rough  angle  measurer.  In 
Measuring  Implements  of  Long  Ago  Mr.  W.  E.  Stark 
describes  some  ancient  forms  of  such  instruments. 

In  finding  heights  the  angle  of  elevation  is  ^ 

used.     If  B  is  the  top  of  a  tower  (Fig.  339),  the 

Z  CAB  is  the  angle  of  elevation.     Notice  that  j^— ^, 

line  AC  is  horizontal.  Fig.  339 

It  is  suggested  that  after  solving  the  following  exercises 
the  pupil  apply  his  knowledge  to  some  practical  problems 
of  his  own  devising. 

Ex.  1.  A  tower  stands  on  level  ground.  The  angle  of  elevation 
of  its  top  at  a  point  160  ft.  from  its  base  is  43°.  Find  the  height 
of  the  tower. 

Ex.  2.  What  is  the  angle  of  elevation  of  the  sun  if  a  tree 
32  ft.  high  casts  a  shadow  18  ft.  long? 

Ex.  3.  What  is  the  height  of  a  balloon  if  its  angle  of  elevation 
is  19°  when  seen  from  a  place  10  miles  from  a  point  directly 
below  it? 

Ex.  4.  The  length  of  a  string  attached  to  a  kite  is  300  ft. 
Find  the  height  of  the  kite  if  its  angle  of  elevation  is  56°. 

Note.  Ex.  4,  of  course,  assumes  that  the  string  is  straight,  which 
is  never  really  true.     Will  the  height  found  be  too  i/rea,t  or  too  small? 

Ex.  5.  A  perpendicular  cliff  650  ft.  high  subtends  an  angle 
of  elevation  of  48°.     How  far  is  it  to  the  base  of  the  cliff  ?    . 

Ex.  6.     Show   how  to  measure  an  angle  A 
indirectly   without   the   use   of   a  protractor  by 
measuring    the    segments   marked   on   Fig.    340. 
If   ZA  is  given,  how   is   the   remainder   of    the^.- 
figure  constructed?  Fig.  340 


RATIO  AND  PROPORTION  197 

SUMMARY  AND   SUPPLEMENTARY  EXERCISES 
229.  SUMMARY  OF  IMPORTANT  POINTS  IN  CHAPTER  IX 

A.  Tests. 

I.  To  prove  two  products  equal,  use  the  factors  of  one 
as  the  extremes  and  the  factors  of  the  others  as 
the  means  of  a  proportion  and  prove  the  ratios 
equal  (Th.  91  and  §211). 
II.  To  prove  two  ratios  equal,  look  for 

a.  Two  similar  triangles  (§211  and  Ths.  101  and 

102). 

b.  Two  transversals  cut  by  three  parallels  (§211 

and  Th.  98). 

c.  A  line  parallel  to  the  base  of  a  triangle  (§211 

and  Th.  99). 

d.  Two  ratios  equal  to  a  third  ratio  (§211  and 

As.  57). 

B.  Algebraic  equations  indicating  constructions. 


I. 

If  T  =  - ,    ax  =  bc, 

0       X 

or  X 

=  — ,  construct  x, 
a 

a  fourth 

proportional  to 

a,  6, 

and  c 

(§207). 

I. 

If  ^  =  ^,  x'  =  ab, 
X      b 

or  :*;  = 

--  ylab, 

construct  x, 

,  a  mean 

proportional  between 

a  and 

b  (def.  Ex. 

4.  §212; 

§221). 

EXERCISES  INVOLVING  EQUAL  RATIOS 
AND  EQUAL  PRODUCTS 

230.  Note.     Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend. 

1.  Make  a  review  diagram  for  Ths.  102,  105,  and  106. 

2.  In  Fig.  341,  0  is  an  arbitrary  point  in  AB.  ^ 
XY  and  ZW  are  any  two  lines  through  point  0. 
If  ^F  and  BX  are  perpendicular  to   XY,  and'*'^^^^^        ^"^ 


AW  and  BZ  perpendicular  to  ZW,  prove  that        \/ »' 
AY    AW  J    ,^, 

BX  =  -BZ'  .  ^"-'^^ 

14 


198 


PLANE  GEOMETRY 


3.  In  Fig.  342,  A  BCD  is  a  parallelogram 
and  Y  is  an  arbitrary  point  in  DC  extended. 

Prove  that  AO^  =  0X  -OY.  p^^,   342 

4.  Draw  any  triangle  and  its  three  altitudes.  Find  all  possible 
pairs  of  similar  triangles  and  read  the  ratios  of  corresponding  sides. 

5.  In  Fig.  343,  P  is  the  mid-point  of  the  arc  CD. 
PA  and  PB  are  arbitrary  chords  intersecting  chord 
CD  at  X  and  Y  respectively.    Prove  that  PY  -  PB=  j 
PA  •  PX. 

6.  Fig.  344  is  a  square  with  its  diagonals  A  C 
and  BD.  AE  bisects  Z  BAC.  The  other  segments 
are  similarly  drawn.  Find  pairs  of  similar  tri- 
angles and  read  the  ratios  of  corresponding  sides. 

Suggestion.  First  prove  that  AE,  CG,  and  DB 
are  concurrent  and  that  AE  and  CF  are  parallel. 
What  other  segments  must  be  proved  concurrent? 
What  ones  must  be  proved  parallel? 

1 7.  If  two  parallel  lines  are  cut  by  a  pencil  of  rays,  correspond- 
ing segments  on  the  parallels  have  equal  ratios.  Investigate  two 
cases. 

t8.  State  and  prove  the  converse  of  Ex.  7. 

9.  In  Fig.  345,  ABC  is  any  triangle.     AX  =  BY. 
XW  II  BC  and  YZ  \\  AC.     Prove  that  WZ  \\  AB. 

10.  If,  in  Fig.   345,    AX  =  BY,    XW\\CB,   and  / 

WZ  II  ^B,  is  FZ  ll^C?     Give  proof.  ^    ^ 


Y 

Fig.  345 

11.  The  common  tangent  to  two  circles  divides  the  segment 
joining  the  centers  into  segments  that  have  the  same  ratio  as  the 
radii.     Investigate  different  cases. 

•      12.  In  Fig.  346,  CO  bisects  ABC  A.     Prove 

a       r  -P 

that  -7-  =  —  by  prolonging  i5C  so  that  CD  =  CA 

and  joining  AD. 

13.  If,  in   Fig.  346, 
bisects  ZBCA. 


—  J-,  prove  that  CO 


B      >■       O    ^  A 

Fig.  346 


RATIO  AND   PROPORTION 


199 


14.  In  Fig.  347,  CO  bisects  the  exterior  /.ACE. 
—  =  —  by  making  CD  =  CA  and  joining  AD. 
BO  =  r  and  AO  =  s. 

15.  If,  in  Fig.  347,  j  =  —,  prove  that  CO 
bisects  Z.ACE. 


Prove  that 


Fig.  347 


16.  In  Fig.  348,  AB  is  a  diameter  of  OO,  CB  and 
AD  are  tangents  at  the  ends  of  the  diameter.  If 
AC  and  BD  intersect  on  the  circle  at  E,  prove  that 


AB 


BC  '  AD. 


17.  In  Fig.  349,  ABC  is  an  isosceles  triangle 
inscribed  in  OO.  Any  line  is  drawn  from  C  cut- 
ting  AB  at  E  and   the   circle   at  D.     Prove   that 

AC^  =CD'CE. 

18.  Investigate  the  case,  Ex.  17,  in  which  CE 
cuts  AB  extended. 


19.  If  one  of  the  parallel  sides  of  a  trapezoid  is  double  the 
other,  the  diagonals  trisect  each  other. 

20.  In  Fig.  350,  A  BC  is  any  triangle  inscribed  in 
the  circle.  •  CQ  bisects  ZC.  Prove  that  CA  •  CB  = 
CP  •  CQ.  A\ 

21.  If  two  chords  intersect  within  a  circle  so  that 
one  of  them  is  bisected  by  the  other,  half  of  the  first 
chord  is  a  mean  proportional  between  the  segments 
of  the  second  chord. 

22.  Use  Ex.  21  to  construct  a  mean  proportional  to  two  given 
segments. 

EXERCISES  INVOLVING  THE  PYTHAGOREAN  THEOREM 

231.  1.  Make  a  review  diagram  for  Th.  110. 

2.  The  radius  of  a  circle  is  48  ft.  Find  the  distance  between 
two  chords  which  are  72  ft.  and  36  ft.  respectively.  What  two 
cases  are  possible? 


200 


PLANE  GEOMETRY 


3.  In  Fig.  351,  ED  is  a  perpendicular  bisector  of  the  chord  AB. 
In  each  case  given  below  construct  the  figure  to  scale  from 
the  data  given.  Compute  the  lengths  of  the  segments 
required  and  verify  your  results  by  measurement. 

a.  CE  =  i,  AB  =  20.  Find  AE,EO,  and  AD. 

b.  AB  =  S6,  £0  =  30.  Find  EC,  ^£,  and  ^Z>. 

c.  AE  =  26,  CE=  10.  Find  AB,  EO,  and  A D. 

d.  AE  =  Q1,  AB  =  120.  Find  CE,  ED,  and  AD. 

4.  The  radius  of  a  circle  is  12  in.  Find  the  length  of  a  tangent 
drawn  from  a  point  13  inches  from  the  center. 

5.  In  Fig.  352,  AX  and  BX  are  tangent  to  OO  from  point  X. 
AB  is  the  chord  joining  the  points  of  contact.  In  each  case  given 
below  construct  the  figure  to  scale  from  the  data 
given.  Compute  the  lengths  of  the  segments  re- 
quired and  verify  your  results  by  measurement. 

a.  AX  =  6,  OX  =  10.  Find^Oand^^. 

b.  AX  =  40,  A0  =  9.  Find  OZ  and  ^5. 

c.  A0=15,  0X  =  S9.  Find  ^Z  and  ^5. 

d.  A0  =  5,  AB  =  8.  Find  ^X  and  OX. 

6.  The  radii  of  two  concentric  circles  are  9  and  15  respectively. 
Find  the  length  of  a  chord  of  the  outer  circle  which  is  a  tangent 
of  the  inner. 

7.  In  the  middle  of  a  pond  10  ft.  square  grew  a  reed.  The  reed 
projected  one  foot  above  the  surface  of  the  water.  When  blown 
aside  by  the  wind,  its  top  reached  to  the  mid-point  of  a  side  of 
the  pond.     How  deep  was  the  pond?     (An  old  Chinese  problem.) 

8.  The  length  of  the  common  chord  of  two  intersecting  circles 
is  16  in.,  the  radii  are  10  and  17  in.  respectively.  Find  the  dis- 
tance between  the  centers. 

9.  The  span  of  a  circular  arch  is  120  ft.  If  the  radius  of  the 
circle  of  which  it  is  a  part  is  720  ft.,  find  the  height  of  the  middle 
of  the  arch. 

10.  Find  the  altitude  of  an  isosceles  trapezoid  if  the  parallel 
sides  are  40  in.  and  58  in.  respectively  and  the  non-parallel  sides 
are  41  in. 


RATIO  AND  PROPORTION  201 

11.  Two  parallel  chords  in  a  circle  are  one  inch  apart.  Find  the 
radius  of  the  circle  if  the  chords  are  8  and  6  inches  long  respectively. . 
Is  there  more  than  one  solution  of  this  problem? 

Suggestion.     Find  two  expressions  each  equal  to  the  square  of  the 
radius  of  the  circle  and  form  an  equation.     Solve  the  equation. 

1 2.  Divide  a  given  segment  in  the  ratio  of  1  to  V2. 

13.  Fig.  353  shows  an  isosceles  right  triangle. 


XY  is  parallel  to  ^C  and  so  constructed  as  to  equal  "*   jf 
CY.     Show  how  to  construct   XY  and   find   the 
ratio  of  CY  to  YO  (see  §86,  Ex.  12). 

14.  Fig.  354  shows  an  isosceles  right  triangle 
with  its  inscribed  circle.  Find  the  ratio  of  CX  to 
XO  and  hence  the  length  of  XO  ii  AB  =  ^. 

15.  In  Fig.  355  the  arcs  AC  and  BC  are  drawn 
with  AB  a.s  radius  and  B  and  A  as  centers  respec- 
tively.   The  circle  O  is  tangent  to  AC,  CB,  and 

to  the  semicircle.      If  AB  =  s  and  the  radius  of    i d 

OO  is  r,  find  r  in  terms  of  5  and  construct  the  ^^^-  ^^^ 

figure. 


A  church  window 
design 


Suggestion.     OB=s-r,    0D  =  }4s-\-r,    DB  =  ^s,    OB^  =  0^ -^-Dl^ - 
Make  the  substitutions  and  solve  the  equation. 

16.  Find  the  shortest  path  that  an  insect  can  take  (without 
flying)  from  one  corner  of  a  room  to  the  diagonally  opposite  corner 
if  the  room  is  15  ft.  long,  12  ft.  wide,  and  10  ft.  high. 

17.  Choose  two  points,  A  and  B,  upon  a  given  straight  line,  and 
two  other  points,  C  and  Z>,  upon  a  straight  line  perpendicular  to 
AB.  Prove  that  the  hypotenuse  of  a  right  triangle  whose  legs 
are  equal  to  ^C  and  BD  is  equal  to  the  hypotenuse  of  a  right 
triangle  whose  legs  are  equal  to  ^Z>  and  BC. — College  Entrance 
Examination  Board,  Plane  Geometry  Examination,  1910. 

18.  A  BCD  is  a  rhombus  with  A  and  C  as  opposite  vertices. 
0  is  a  point  within  the  rhombus  such  that  OB  =  OD.  Prove 
that  A ,  O,  and  C  are  on  the  same  straight  line,  and  that  OA  •  OC  = 

2       2 

AB  —OB  . — College  Entrance  Examination  Board,  Plane  Geom- 
etry Examination,  1916. 


202 


PLANE  GEOMETRY 


19.  A  sloping  embankment  rises  from  a  level  field.  One  end 
of  a  prop,  20  ft.  long,  rests  on  the  ground  16  ft.  from  the  foot  of 
the  embankment,  and  the  other  end  rests  9  ft.  up  the  embankment, 
measured  along  its  sloping  side.  How  high  is  the  upper  end  of 
the  prop  above  the  level  field?  Result  in  feet  to  one  decimal. — 
College  Entrance  Examination  Board,  Plane  Geometry  Exami- 
nation, 1914. 


20.  If,  in  Fig.  356,  AB  =  AC  =  \,  show  that  ^ 
BC=  yJ2.     If  AD  =  BC,  what  is  the  length  of 
BD}     If  AE  =  BD,  what  is  the  length  of  BE? 
Show  how  the  figure  may  be  continued  so  as  to 
construct  segments  equal  to  V5,  V6,  V7,  etc. 


c 
Fig. 


D  E 

356 


21.  In  Fig.  357,  A  BCD  is  a  square.    D  is  the  center  and  DB 
is  the  radius  for  the  arc  BE ;  A  is  the  center     j>  c   e      g 

and  ^£  is  the  radius  for  the  arc  EF;  D  is  the 
center  and  DF  is  the  radius  for  the  arc  FG, 
etc.  Find  the  length  of  DE,  AF,  DG,  AH, 
etc.,  if  AB  =  1. 


Fig. 


F 

357 


EXERCISES  INVOLVING  THE  TRIGONOMETRIC  TABLES 

232.  1.  Find  the  legs  and  the  altitude  of  an  isosceles  triangle 
if  the  base  is  24  and  each  acute  angle  is  49°. 


2.  One  of  the  equal  sides  of  an  isosceles  triangle  is  45  and  each 
base  angle  is  68°.     Find  the  base  and  the  altitude. 


3.  The  distance  across  a  stream  may  be  found 
as  follows  (Fig.  358):  Lay  off  AC  ±  AB, 
extending  AC  to  some  point  from  which  B  is  visible. 
Measure  AC  and  angle  C.  Find  ^15  if  ^C  =  300 
ft.  and  ZC  =  56°. 


4.  Find  the  distance  AC  across  a  pond  as  shown  in 
Fig.  359.  CB  is  perpendicular  to  AC,  Z5  =  34°, 
C5  =  165ft. 


Fig.  359 


RATIO  AND   PROPORTION  203 

5.  A  chord  of  a  circle  is  8  in.  It  subtends  at  the  center  of  a 
circle  an  angle  of  36°.  Find  the  radius  of  the  circle  and  the  dis- 
tance of  the  chord  from  the  center. 

6.  An  angle  at  the  center  of  a  circle  of  radius  6  ft.  is  40°.  Find 
the  length  of  the  subtended  chord  and  the  distance  of  the  chord 
from  the  center  of  the  circle. 

7.  Prove  that  in  any  triangle  -7-=  ~. — b  (see  Fig.  360). 

0       sin  £> 

Suggestion.     From  the  figure  find  sin  A  and  sin  B  and 
divide  one  equation  by  the  other.  ^ 

8.  Find  the  sides  of  a  triangle  if  Z^=42°,  ZB  = 
65°,  and  AB  =  8.     Use  the  formula  obtained  in  Ex.  7. 

9.  Find  the  sides  of  a  triangle  if  Z^=68°,  ZC  =  49°, 
and  a  =  25. 

10.  Two  observers  IX  miles  apart  observe  at  the  same  moment 
the  altitude  of  the  base  of  a  thundercloud  that  is  between  them. 
If  the  angles  are  42°  and  61°,  how  high  was  the  cloud? 

Note.  Ex.  10  illustrates  a  method  actually  used  by  weather  bureau 
men.  The  two  observers  are  in  telephonic  communication,  select  some 
singular  part  of  the  cloud  that  neither  can  fail  to  recognize,  and  take 
the  observation  at  a  stated  time  by  the  watch. 


MISCELLANEOUS  EXERCISES 

233.     Note.     Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend. 

1.  In  drawing  a  certain  map  all  segments  are  reduced  in  the 

ratio  .       What    lengths   will   represent    the   sides    of    a 

100  000 

coimty  which  is  a  rectangle  25  miles  long  and  18  miles  wide? 

2.  In  Fig.  301,  ABC  is  an  iscsceles  triangle.  ^ 


CE-=EB  =  CF  =  FA.    FG,  CO,  and  EH  are  A.  AB       f^^'^^k 
from  F,  C,  and  £  respectively.     If  ^  =  -^,  what '^     g     o     h     b 


CO     '^  ^-^ — '-     ^- 

BO      8'  Fig.  361 


is  the  value  of  II?  '  ^'"""^  IS^'''^'' 


204 


PLANE  GEOMETRY 


3.  In  Fig.  362,  A  BCD  is  a  square.  AK  =  BF== 
CG  =  DH,  XY  and  ZW  are  medians.  GN  is  par- 
allel to  AD  from  G.  li  DG=H  DC,  what  is  the 
ratio  of  HM  to  MK  ?  If  HM  =  %  HK,  what  is  the 
ratio  of  DG  to  DC? 


D       G   Z  C 

7         I  / 


Fig.  362 

From  a  Pompeian 

mosaic 


4.  Restate  Ths.  103  and  104  so  that  two 
ratios,  rather  than  two  products,  are  to  be  proved 
equal.  In  Th.  103  the  segments  of  the  chords,  and  in  Th.  104  the 
segments  of  the  secants,  are  inversely  or  reciprocally  proportional . 
Why? 

5.  In  Fig.  363,  0)0  and  X  intersect  at  points 
A  and  B.  AD  and  AC  are  tangent  to  (DX  and 
0  respectively  at  point  A.  Prove  that  AB  is  a 
mean  proportional  between  DB  and  BC. 

6.  Fig.  364  shows  the  outline  of  a  roof  truss.  k^^tv.  h 

ABC  is  an  isosceles  triangle.     The  equal  sides     ^S<1\ /fS^ 

are  each  divided  into  3  equal  parts.     CD,  EF,  a        f   d  k       b 

and  HK  are ±  AB.    If  AC=35,  and  CD=^  AB,  Fig.  364 

From  a  roof  truss 
design 


find  the  length  of  AB,  CD,  EF,  and  ED. 


7.  In  Fig.  365,  ABC  is  any  inscribed  triangle. 
Z1=Z2.    Frove  that  AC'  RB  =  CR'  AT  and  that   a^ 
AC'TB  =  CT-AR, 

Fig.  365 

8.  Two  tangents  each  24  in.  long  are  drawn  from  the  same 
point  to.  a  circle  of  radius  7  in.  Find  the  length  of  the  chord 
joining  the  points  of  contact. 

t9.  A  perpendicular  is  drawn  from  the  vertex  of  the  right 
angle  of  a  right  triangle  to  the  hypotenuse.  Prove  that  the  ratio 
of  the  squares  of  the  legs  equals  the  ratio  of  the  adjacent  segments 
of  the  hypotenuse.  c 

10.  In  Fig.  366,  ABC  is  any  inscribed  triangle. 
CN  is  perpendicular  to  AB  and  CD  is  a  diameter. 
Prove  that  (1)  CB  -  AN  =  CN  •  DB;  (2)  AC  -  NB  = 
CN'  AD;  (3)  AC-  CB  =  CD'CN. 


RATIO  AND  PROPORTION 


205 


11.  Fig.  367  shows  a  diagram  of  the  roof  of  a  barn.   DE  A_  AF. 
The  rafters  AD  and   FD   make   equal    angles  ^ 
with   DE.     If  ^£  =  26   ft.,    AD  =  ZO   ft.,   and 
Z?C  =  13  ft.,  find  CB.     CB  is  ±  ^F. 

Pig.  367 

12.  Find  by  geometry  two  segments  whose  sum  is  equal  to 
a  segment  5  cm.  long  and  whose  ratio  is  3:7.     Find  by  algebra 

a      3 

the  value  of  a  and  b  if  a-{-b  =  5  and  -r=='^' 

0      7 

13.  Find  by  geometry  two  segments  whose  difference  is  1.5 
cm.,  and  whose  ratio  is  5  to  8.    Find  by  algebra  the  value  of 


a  and  6  if  a  ^  6  = 


1.5  and -r= -3- 
0      8 


^  14.  In  Fig.  368,  ABC  is  an  isosceles  triangle. 
CE=y3  CB,  CF^VsCA.  FG,  CO,  and  EH  are  ± 
AB  from  F,  C,  and  E  respectively.  Prove  that  if 
EFGH  is  a  square  CO=OB.     What  must  be  the 


A 


G    O    H  B 

Fig.  368 

ratio  of  CO  to  OB  if  CE  =  )4  CB  and  EFGH  is   From  a  roof  truss 
a  square?     Construct  the  figure  in  each  case.  design 


15.  Inscribe  a  square  in  a  given  triangle. 

Suggestion.    The  construction  is  suggested  in  Fig. 

)9.     To  prove  that  HG  =  HE,  prove  £^  =  ^. 

HG    HE  Fig.  369 

16.  Solve  Ex.  15  by  the  construction  shown  in  Fig.  370. 


A 


Fig.  370 


Fig.  371 


Fig.  372 


17.  Show  how  to  inscribe  a  square  in  a  sector  (1)  as  shown  in 
Fig.  371 ;    (2)  as  shown  in  Fig.  372. 


18.  Inscribe  a  square  in  a  given  semicircle. 

Suggestion.  Solve  by  at  least  two  methods;  use  the 
method  employed  for  Fig.  372;  also  that  suggested  by 
Fig.  373.  A 


^ 


c   K    n 
Pig.  373 


206 


PLANE  GEOMETRY 


19.  Let  ABC  be  a  triangle  with  a  right  angle  at  C.  Draw 
CD  and  CE  equally  inclined  to  CB,  and  meeting  AB  (or  AB  pro- 
longed) in  D  and  E  respectively.  Let  M  be  the  mid-point  of  AB. 
Prove  that  MB  is  a  mean  proportional  between  MD  and  ME. — 
College  Entrance  Examination  Board,  Plane  Geometry  Examina- 
tion, 1910. 

20.  The  distance  between  two  parallel  chords  on  the  same  side 
of  the  center  of  a  given  circle  is  6  cm.  If  the  chords  are  36  cm. 
and  48  cm.  respectively,  find  the  radius  of  the  circle.  What  would 
be  the  distance  between  the  chords  if  they  were  on  opposite  sides 
of  the  center? 

21.  If  two  circles  are  tangent  externally  and  a  segment  is 
drawn  through  the  point  of  contact  terminated  by  the  circles, 
the  chords  intercepted  in  the  two  circles  have  the  same  ratio  as 
the  radii. 

22.  Fig.  374  represents  a  gable  over  an  equilateral  Gothic  arch. 
The  arcs  CA  and  CB  are  drawn  with  B  and   A   as 

centers  and  A  B  as  radius.  The  sides  of  the  gable  DE 
and  DF  are  tangent  to  the  sides  of  the  arch  from  a 
point  in  the  common  chord  GC  extended. 

a.  Prove  that  DE  =  DF. 

b.  Construct  the  figure  so  that  AB  =  6  cm.  and 
DC  =  4cm. 

c.  Construct  the  figure  so  that  AB  =  Q  cm.  and 
DE=  10  cm. 

d.  Construct  the  figure  so  that  AB  =  Q  cm.  and   Z£Z)F  =  30' 

23.  Prove  the  Pythagorean  theorem  by  means 
of  Fig.  375. 

Suggestion.     Let  a  represent  the  distance  OB,  b  repre- 
sent the  radius  of  O  O,  and  c  represent  the  half  chord 
perpendicular  to  ^  C  at  B. 
or  &2_a2  =  c2. 


Fig.  374 


Prove  that  {b—a)  {b+a) 


24.  Construct  a  segment  x  so  that  x  =  ^  ^ab;  x 

2^2 
x  =  — ,  where  a,  b,  and  c  are  given  segments. 


RATIO  AND  PROPORTION  207 

25.  Show  that  the  following  construction  will  give  graphically 
the  solution  of  the  equation  x^  —  2x  =  24: 

Construct  a  circle  whose  diameter  is  2.  _At  any  point  on  the 
circle  construct  a  tangent  whose  length  is  V24.  From  the  end  of 
the  tangent  draw  a  secant  which  passes  through  the  center  of  the 
circle.  The  entire  secant  will  be  one  of  the  roots  of  the  given 
equation. 

26.  Ovals  are  of  frequent  use  in  landscape  gardening  and  other 
branches  of  engineering.  Agreeable  ovals  may  be  laid  out  as 
follows  (Fig.  376):  Let  AB  be  the  total 
length  of  the  desired  oval.  Lay  ofi  on  AB 
two  equal  intersecting  circles  with  any  radius. 
Draw  the  common  chord  CC  and  extend  CC\ 
making  CD  equal  to  a  diameter  of  the  circles. 
Through  D  draw  lines  through  the  centers  of 
the  circles  O  and  O'  intersecting  the  circles  at 
X  and  Y.  With  DX  as  radius  draw  XY:  In  a  similar  manner 
draw  ZW .  Show  that  the  circles  are  tangent  at  X .  If  ^0'  =  HAB 
and  AB==lo  ft.,  find  DD' . 

27.  In  Fig.  377,  ABC  is  any  triangle  inscribed  in 
OO.  CX  is  the  altitude  to  AB,  A  Y  is  the  altitude 
to  CB.  OK  is  the  perpendicular  from  the  center  of 
the  circumscribed  circle  to  A  B.  H  is  the  intersection 
of  the  altitudes.     Prove  OK  =  14  CH.  Fig.  377 

28.  Find  the  locus  of  points  from  which  the  distances  to  two 
given  intersecting  lines  are  in  a  given  ratio. 

29.  Investigate  the  case,  Ex.  28,  in  which  the  two  given  lines 
are  parallel. 

30.  Two  equal  circles  intersect  in  such  a  manner  that  the 
common  chord  is  equal  to  the  segment  joining  the  centers.  If  the 
common  chord  is  2  in.,  find  the  radius  of  the  circles  and  the  width 
of  the  oval  formed.  If  the  radius  is  3  in.,  find  the  common  chord 
and  the  width  of  the  oval  formed. 

31.  Find  the  hypotenuse  of  a  right  triangle  if  the  legs  are  (1)  n 
and  K(""~l);  (2) »  and(>^«)2— 1.  Verify  in  each  case  by 
substituting  numbers  for  n. 


CHAPTER  X 

Area  and  Equivalence 

INTRODUCTORY 

MEASURING  SURFACES 

234.  To  measure  the  surface  inclosed  by  the  sides  of  a 
polygon  is  to  find  how  many  times  it  contains  another  sur- 
face chosen  as  a  unit  of  measure. 

The  area  of  a  polygon  is  the  measure  number  of  the 
surface  of  the  polygon. 

It  is  the  common  practice  to  use  as  a  unit  of  surface  a 
square  whose  side  is  a  unit  of  length.  Thus  if  the  unit  of 
length  is  an  inch,  the  unit  of  surface  is  a  square  whose  side 
is  an  inch  and  is  called  a  square  inch. 

While  any  segment  may  be  used  as  a  unit  of  length  with 
its  corresponding  unit  of  surface,  it  is  most  convenient  prac- 
tically to  use  one  of  the  recognized  standard  units  of  length, 
such  as  the  inch,  foot,  yard,  mile,  tenth  of  an  inch,  centi- 
meter, etc.,  with  their  corresponding  units  of  surface — 
square  inch,  square  mile,  etc.  The  particular  unit  chosen 
depends  upon  the  surface  to  be  measured. 


EQUIVALENT  POLYGONS 


235.  Ex.  1.  In  Fig.  378,  AABC  is  isosceles 
and  CD  is  perpendicular  from  C  to  AB. 
Show  that  A  a  and  h  are  congruent.  Draw 
the  AABC;  cut  out  Aa  and  b  and  place 
them  together  so  that  AD  coincides  with  DB 
so  as  to  form  AXYZ.  Are  AABC  and  XYZ 
congruent?  Do  they  cover  the  same  extent 
of  surface? 


208 


AREA  AND  EQUIVALENCE  209 

Two  polyp^ons  that  cover  the  same  extent  of  surface  are 
called  equivalent  polygons.  The  symbol  (  =  )  is  used  for 
equivalence.     Since  area  is  the  measure  of  surface, 

1.  Equivalent  polygons  have  equal  areas. 

2.  Polygons  with  equal  areas  are  equivalent. 

Since  congruent  polygons  can  be  made  to  coincide,  they 
may  be  made  to  cover  the  same  surface  and  are  equivalent. 
Congruent  polygons  are  the  simplest  examples  of  equivalent 
polygons. 

It  does  not  follow  that  equivalent  polygons  are  always 
congruent.  The  following  exercise  gives  illustrations  of 
polygons  that  may  be  made  to  cover  the  same  surface  but 
are  not  necessarily  congruent. 

Ex.  2.  Construct  two  congruent  right  triangles  that  are  not 
isosceles.  Cut  them  out  and  place  them  together  in  different 
positions  so  as  to  form  two  isosceles  triangles,  a  rectangle,  a  kite, 
two  oblique  parallelograms,  and  other  polygons,  all  of  which  are 
equivalent.  Make  careful  drawings  of  these  figures.  The  use  of 
cross-section  paper  is  suggested. 

236.  It  is  evident  that  combinations  of  congruent  polygons 
will  give  polygons  that  are  equivalent  but  not  necessarily 
congruent.     The  following  definitions  are  necessary: 

If  two  polygons  are  so  placed  that  a  side  of  one  falls  upon 
a  side  of  the  other,  but  neither  polygon  overlaps  the  other, 
the  polygon  inclosed  by  the  entire  perimeter  is  the  sum  of 
the  two  polygons. 

If  one  polygon  is  placed  entirely  within  another,  the  space 
between  the  perimeters  is  the  difference  between  the 
polygons. 

A  polygon  is  bisected  by  a  segment  if  the  segment  divides 
it  into  two  equivalent  parts;  for  example,  the  diagonal 
of  a  parallelogram  bisects  the  parallelogram.  Similarly,  a 
polygon  may  be  trisected  or  may  be  divided  into  any 
number  of  equivalent  parts. 


210  PLANE  GEOMETRY 

One  figure  is  transformed  into  a  second  if  the  second  is 

equivalent  to  the  first. 

237.  The  following  assumptions  will  be  used  in  the  discus- 
sion of  equivalent  figures: 

As.  59.  If  equivalent  polygons  are  added  to  equivalent 
polygons,  the  results  are  equivalent  polygons. 

As.  60.  If  equivalent  polygons  are  subtracted  from  equiv- 
alent polygons,  the  results  are  equivalent  polygons. 

As.  61.  If  equivalent  polygons  are  divided  into  the  same 
number  of  equivalent  polygons,  each  part  of  one  is  equivalent 
to  any  part  of  the  other. 

As.  62.  Polygons  equivalent  to  the  same  polygon  or  to 
equivalent  polygons  are  equivalent. 

238.  We  have  then  the  following  preliminary  test  for 
equivalent  polygons : 

I.  To  prove  two  polygons  equivalent,  prove  that  they  are 
made  up  of  parts  congruent  in  pairs. 

II.  To  construct  two  polygons  equivalent,  construct  them 
of  parts  congruent  in  pairs.  y 

Ex.  1.     Given  AABC  with  D  the  mid-  ^/..[\e._..^f 

point  of  AC.     DF\\AB  and  BF  ||  AC.     Prove  /  \  "y''' 

AABC=OJABFD  (Fig.  379).  a^ i/ 

Fig.  379 
A  nalysis: 

I.  To  prove    AABC=OJABFD,   prove  that  they 
are  the  sums  of  congruent  parts. 

II.  .-.prove    ABED  +  Al=ABED  +  All. 

III.  .*.  prove  AI  ^  AIL 


Ex.  2.     Transform   a   trapezoid    into    a  paral 
lelogram  (Fig.  380). 


A      E  B 

Fig.  380 
A  nalysis: 

To  transform  A  BCD  into  a  parallelogram,  construct  the  parts  of 
the  parallelogram  congruent  to  the  parts  of  A  BCD. 


AREA  AND  EQUIVALENCE 


211 


Ex.  3.     Transform  a  triangle  into  a  rectangle. 

Ex.  4.     Transform  a  trapezoid  into  a  rectangle. 

Ex.  5.  Any  segment  through  the  intersection  of  the  diago- 
nals of  a  parallelogram  and  terminated  by  the  sides  divides  the 
parallelogram  into  two  equivalent  parts. 

Ex.  6.  Show  how  to  bisect  a  parallelogram  by  a  line  (1)  perpen- 
dicular to  the  base  and  (2)  parallel  to  the  base. 


239. 


MEASUREMENT   OF   POLYGONS 

FUNDAMENTAL   ASSUMPTION 

As.  63.     The  number  of  units  of  area  in  a  rectangle 


is  equal  to  the  product  of  the  number  of  units  of  length  in 
the  base  and  altitude. 

If  S  represents  the  area  of  a  rectangle,  b  the  length  of  its 
base,  and  a  the  length  of  its  altitude,  As.  63  may  be  stated 
as  a  formula,  S  =  ab. 

The  assumption  will  be  discussed  under  two  heads : 
A.  When  the  sides  of  the  rectangle  are  both  commensurable 
with  a  given  unit  of  length. 

In  this  case  the  unit  of  length  can  be  applied  an  integral 
number  of  times  to  both  the  base  and  the  altitude  of  the 
rectangle.  The  assumption  is  evident  at  once.  The  unit 
of  length  chosen  may  be  contained  in  the  base  m  times  and 
in  the  altitude  n  times,  if  m  and  n  are  whole  numbers.  By 
drawing  the  proper  lines  the  rectangle  may  be  divided  into 
n  rows  with  m  unit  squares  in  a  row. 

Illustration  1-  Suppose  the  unit  chosen  is  a  square  centimeter  and 
the  rectangle  is  3  cm.  long  and  2  cm.  wide  (Fig.  381).     The  unit  of 

length  is  contained  in  the  base  3  times  and  p, , , ,0 

in  the  altitude  2  times.  By  drawing  seg- 
ments through  the  points  of  division  par- 
allel to  the  sides,  the  rectangle  is  divided  ~ 
into  2  rows  with  3  sq.  cm.  in  a  row,  or  into 
2X3  sq.  cm.,  or  6  sq.  cm.  The  measure 
number  of  the  surface  is  6;  the  area  of  the  a" 
surface  is  6  sq.  cm. 


3  cm. 

Fig.  381 


212  PLANE   GEOMETRY 

Sometimes  one  or  both  of  the  sides  of  the  given  rectangle 
are  not  exactly  divisible  by  the  unit  chosen,  but  are  divisible 
by  some  aliquot  part  of  this  unit.  In  this  case  this  part 
of  the  chosen  unit  may  be  taken  as  a  new  unit  of  length, 
and  a  square  whose  side  is  this  new  linear  unit  may  be  con- 
sidered as  the  unit  of  area.  The  assumption  is  then  evident 
as  above. 

Illustration   2.     Suppose   the   unit   chosen   is   a   square   inch.     In 
Fig.  382  one  inch  is  not  contained  exactly  in  either  AB  or  AC.     One 
quarter-inch  is,  however,  exactly  contained 
in  both  AB  and  A C.   One  quarter-inch  may 
be  used  as  a  convenient  linear  unit.     The 
measure  number  of  ^5  is  5  and  of  A  C  is  3. 
A  square  quarter-inch  may  be  considered  as 
the  unit  of  area.     By  drawing  the  proper 
lines  we  can  show  that  the  rectangle  con-^ 
sists  of  3  rows  with  5  units  of  surface  in  a  Fig.  382 

row.     The  measure  number  of  the  area  is  15. 

In  this  case  we  may  express  the  unit,  the  measure  of  the  sides,  and 
the  area  in  fractional  terms  of  a  larger  unit. 

The  unit  one  quarter-inch  is  M  in. 

The  measure  of  AB,  5  quarter-inches,  is  ^  in. 

The  measure  of  ^C,  3  quarter-inches,  is  ^  in. 

The  area,  15  square  quarter-inches,  is  i^e  sq.  in. 

B.  When  one  or  both  sides  of  the  rectangle  are  incom- 
mensurable with  the  chosen  unit. 

In  this  case  it  is  not  possible  to  measure  one  side  or  perhaps 
both  sides  of  the  rectangle  in  integral  or  fractional  terms  of 
the  chosen  unit.  Since  the  ratio  of  two  incommensurable 
segments  is  an  irrational  number  (§  198),  these  sides  may  be 
expressed  in  irrational  terms  of  the  chosen  unit,  and  are 
measured  approximately.  From  these  approximate  lengths 
an  approximate  area  is  computed  by  the  rule  contained  in 
the  assumption.  We  have  seen  (§194)  that  by  subdividing 
the  unit  of  length  we  can  obtain  approximate  measures  for 
the  sides  of  the  rectangle  that  are  as  close  as  we  choose  to 
make  them.  It  is  evident  that  the  approximation  for  the 
area  may  also  be  m.ade  as  close  as  we  choose  to  make  it. 


AREA  AND  EQUIVALENCE  213 

Illustration  3.  Suppose  the  unit  chosen  is  one  square  centimeter. 
In  Fig.  3S3,AB  is  2  cm.  and  is  commensurable  with  the  unit;  AC 
is  equal  to  the  diagonal  of  a  square  whose  side 
is  2  cm.  and  is  incommensurable  with  the  unit. 
The  length  oi  AC  cannot  be  expressed  in  integral 
or  fractional  terms  of  the  unit.  We  know,  how- 
ever, that  we  can  express  i4  C  as  2  V2  cm.  Accord- 
ing to  our_  assumption,,  therefore,  the  area  of  R 
is  2X2  V2"or  4  V2'sq.  cm.  While  the  length  of 
AC  cannot  be  expressed  exactly,  an  approximate 
length  can  be  found  for  it.  From  this  approximate 
length  an  approximate  area  can  be  found  for  R.  Fig.  383 

These  approximate  values  can  be  made  as  close  as_we  choose. 

Suppose  that  1.4  is  taken  as  the  approximate  V2,  then  2X1 .4  cm., 
or  2.8  cm.,  is  the  approximate  length  of  AC,  and  2X2.8  sq.  cm.,  or 
5.6  sq.  cm.,  is  the  approximate  area  of  R.  In  Fig.  383,  the  area  of  the 
rectangle  ABGH  represents  this  approximate  area  of  R. 

Suppose,  again,  that  1.41  is  taken  as  the  approximate  V2,  then 
2X 1 . 41  cm.,  or  2 .82  cm.,  is  the  approximate  length  of  A  C,  and  2X2 . 82 
sq.  cm.,  or  5.64  sq.  cm.,  is  the  approximate  area  of  R.  Although  this 
approximation  cannot  be  represented  on  the  figure,  we  know  that 
it  is  closer  than  the  other  but  a  little  less  than  the  area  of  R. 

In  this  case  the  following  points  should  be  noted : 

1.  The  side  or  sides  that  are  incommensurable  with  the 
unit  cannot  be  expressed  in  integral  or  fractional  terms  of 
that  unit. 

2.  The  side  or  sides  that  are  incommensurable  with  the 
unit  can  be  expressed  in  irrational  terms  of  that  unit.  In 
illustration  3,  AC  is  expressed  as  2V2  cm. 

3.  It  can  be  proved  that  As.  63  is  true  for  those  cases  in 
which  the  sides  of  the  rectangle  can  be  expressed  only  in 
irrational  tenns  of  the  chosen  unit.  The  proof  is,  however, 
too  difficult  for  this  course. 

4.  Sometimes  when  the  sides  of  the  rectangle  can  be 
expressed  only  by  irrational  numbers,  it  is  possible  to  express 
the  area  of  the  rectangle  by  rational  numbers. 

Illustration  4.  Connect  the  mid-points  of  the  sides  of  a  square  whose 
side  is  4  cm.     Find  one  side  of  the  square  so  formed  and  its  area. 

15 


214 


PLANE  GEOMETRY 


240.  Practical  measurements.  When  it  is  desired  to  use 
the  rule  contained  in  As.  63  to  compute  the  area  of  a  given 
rectangle  from  measurements  actually  made,  an  approximate 
area  only  is  possible. 

We  have  seen,  §  195,  that  the  exact  length  of  a  given 
segment  cannot  be  obtained  in  terms  of  a  unit  chosen  in 
advance.  Since  the  measures  of  the  sides  must  of  necessity 
be  approximate,  the  area  must  of  necessity  be  approximate 
also.  The  approximate  area  may,  however,  be  made  as 
close  as  we  choose  if  only  the  divisions  of  the  scale  are  made 
sufficiently  small. 

EXERCISES  INVOLVING  AREA  OF  RECTANGLES 

241.  1.  Draw  a  rectangle  whose  sides  are  3.4  cm.  and  2.6  cm. 
Find  the  approximate  area  of  this  rectangle  in  inches,  measuring 
(1)  to  the  nearest  inch,  (2)  to  the  nearest  half  inch,  (3)  to  the 
nearest  quarter  inch,  (4)  to  the  nearest  sixteenth  inch.  Make  an 
accurate  drawing  for  each  approximation  and  compare  it  with 
the  given  rectangle. 

2.  Find  the  area  of  a  rectangle  whose  sides  are  SHe  in.  and 
4^2  in. 

3.  If  the  area  of  a  rectangle  is  321^6  sq.  in.  and  one  side  is 
3M  in.,  find  the  other  side. 

4.  Find  the  area  of  a  walk  3  ft.  6  in.  wide  which  completely 
surrounds  a  lot  300  ft.  X  500  ft.  The  dimensions  of  the  lot  are 
taken  on  the  inside  of  the  walk. 


>■' 


7^^" 


Fig.  384 


n 


5.  Fig.  384  shows  forms  of  columns  in  cross-section.  The 
outside  measures  and  the  width  are  the  same  in  each  case.  Find 
the  area  of  each  cross-section  shown. 

Note.     The  area  of  the  cross-section  is  an  important  element  in 
determining  the  strength  of  the  column. 


AREA  AND  EQUIVALENCE  215 

6.  Using  any  three  given  segments  a,  b,  and  c,  construct  a 
rectangle  whose  sides  are  (a+b)  and  c.  Show  how  this  figure 
illustrates  geometrically  the  algebraic  identity  c(a-\-b)=ac+bc. 

Suggestion.  c(a-\-b)=ac-\-bc  may  be  translated  into  geometry: 
The  rectangle  whose  side3  are  c  and  {a-\-b)  may  be  divided  into  two 
rectangles  whose  sides  are  a  and  c,  and  b  and  c,  respectively. 

7.  Illustrate  geometrically  the  identity  c{a  —  b)  =  ac  —  bc. 
Suggestion.     c{a  —  b)=ac  —  bc    may    be    translated    into    geometry: 

The  rectangle  whose  sides  are  c  and  {a  —  b)  may  be  obtained  by  cutting 
a  rectangle  whose  sides  are  b  and  c  from  a  rectangle  whose  sides  are 
a  and  c. 

8.  Using  any  two  given  segments  a  and  b,  construct  a  square 
on  the  segment  {a-\-b)  and  illustrate  geometrically  the  identity 

Suggestion.  Translate  {a-\-bY  =  a'^-\-b'^-\-2ab  into  geometry.  Draw 
the  segments  necessary  to  divide  the  square  on  a-\-b  into  the  required 
parts. 

9.  Illustrate  geometrically  the  identity  {a  —  bY  =  a^-\-b^  —  2ab. 
Suggestion.  Translate  {a  —  bY=a^-\-b'^  —  2ab  into  geometry.  Con- 
struct a  figure  formed  by  adding  a  square  whose  side  is  a  to  a 
square  whose  side  is  b.  Show  how  two  rectangles  may  be  cut  from 
this  so  as  to  leave  the  desired  result.  What  will  be  the  sides  of  these 
rectangles? 

10.  Translate  into  geometry  and  illustrate  by  a  figure: 

a.  {a-\-b)  {d  +  c)  =  ad^bd-\-ac-\-bc. 

b.  {a^-l)^)  =  {a-b)  (a  +  6). 

11.  If  the  length  of  a  rectangle  is  3  ft.  more  than  the  width  and 
the  area  is  70  sq.  ft.,  find  the  dimensions. 

12.  If  the  length  of  a  rectangle  is  2K  times  the  width  and  the 
area  is  360,  find  the  dimensions. 

13.  The  length  of  a  rectangle  is  14  ft.  more  than  its  width.  If 
its  diagonal  is  26  ft.,  find  the  dimensions  and  area. 

14.  Find  the  dimensions  of  a  rectangle  if  its  perimeter  is  38  in. 
and  its  area  84  sq.  in. 

15.  Find  the  dimensions  of  a  rectangle  if  its  diagonal  is  13 
in.  and  its  area  60  sq.  in. 


216  PLANE   GEOMETRY 

MEASUREMENT   OF  THE  PARALLELOGRAM 
242.  Theorem  113.    The  area  of  a  parallelogram  is  the 
product  of  the  base  and  altitude. 


V 

/ 

p 

/ 
/ 

a     / 

I 

b                    B 

Fig.  385 

Hypothesis:    P  is  a  /Z7  with  base  h  and  altitude  a. 
Conclusion:    Area  P  =  ab. 
Analysis  and  construction: 
I.  To  prove  area  P  =  ah,  compare  P  with  a  rectangle,  R, 

that  has  h  for  base  and  a  for  altitude. 
II.   .*.  construct  rectangle  R  with  b  for  base  and  between 
the  parallels  AB  and  CD. 

III.  To  prove  area  P=  area  R,  prove  that  P  and  R  are 

equivalent. 

IV.  .-.  prove  /\ADF  ^  ABCE. 
Proof: 

STATEMENTS  REASONS 

I.    AADF^ABEC.  I.  Let  the  pupil  give  de- 

tails in  full. 
II.    1.  Pis  equivalent  to  i?.        II.  1.  As.  60. 
2.    .-.area  P  =  area  i?.  2.  See  §235. 

III.  Area     R  =  ab,  III.  Why? 

IV.  .-.area  P  =  a6.  IV.  Why? 

Ex.  1.  Prove  Th.  113  for  the  case  in  which  points  E  and 
F  both  fall  on  CD  extended. 

Ex.  2.  Construct  a  parallelogram  whose  sides  are  5  cm.  and  8 
cm.  and  one  of  whose  angles  is  60°.     Find  its  altitude  and  area. 

Ex.  3.  Construct  a  parallelogram  whose  sides  are  6  cm.  and  9.4 
cm.  and  one  of  whose  angles  is  30°.    Find  its  altitude  and  area. 

Ex.  4.  Construct  a  parallelogram  with  sides  4.2  cm.  and  5.9 
cm.  and  one  angle  45°.     Find  its  altitude  and  area. 


AREA  AND  EQUIVALENCE 


217 


MEASUREMENT   OF  THE  TRIANGLES 
243.  Theorem  114.    The  area  of  a  triangle  is  one-half 
the  product  of  the  base  and  altitude. 


Fig.  386 
Hypothesis:    ABC  is  a  A  with  base  h  and  altitude  a. 
Conclusion:    Area  ABC  =  \  ah. 
Analysis  and  construction: 

I.  To  prove  that   area  ABC  =  \  ah,  compare    A  ABC 
with  a  parallelogram  whose  sides  are  AB  and  BC. 

II.    .*.   construct 

Let  the  pupil  give  the  construction. 
Proof: 

STATEMENTS 

1.  A  ABC  is  equivalent  to  J  the  parallelogram. 

2.  Area   EJ=ah. 

3.  .'.  area  A  =  i  a6. 
Let  the  pupil  give  the  reasons. 

Ex.  1.  Find  the  area  of  a  triangle  whose  base  is  6  ft.  3  in.  and 
whose  altitude  is  2  ft.  7  in. 

Ex.  2.  Find  the  altitude  of  a  triangle  if  its  area  is  S%  sq.  ft. 
and  its  base  is  6  ft.  7  in. 

Ex.  3.  Find  the  area  of  an  isosceles  triangle  if  its  base  is  32  ft. 
and  one  leg  is  34  ft. 

Ex.  4.  Find  the  area  of  an  equilateral  triangle  if  each  side 
is  4  in.;  6  in.;  8  in. 

Ex.  5.  The  base  of  a  triangle  is  3  ft.  more  than  its  altitude. 
Find  the  base  and  the  altitude  if  the  area  is  90  sq.  ft. 

Ex.  6.  The  base  of  a  triangle  is  3  times  its  altitude.  Find 
the  base  and  the  altitude  if  the  area  is  336  sq.  ft. 


218 


PLANE   GEOMETRY 


MEASUREMENT  OF  THE  TRAPEZOID 

244.  Theorem  115.     The  area  of  a  trapezoid  is  equal  to 
one-half  the  product  of  the  altitude  and  the  sum  of  the  bases. 


Fig.  387 

Hypothesis:    ABCD  is  a  ZZ\  with  bases  b  and  b'  and  alti- 
tude a. 
Conclusion:    Area  ABCD  =  |  a{b-\-b') . 
Analysis  and  construction: 
I.  To  prove  area  ABCD  =  l  a{b+b'),  divide  ABCD  into 
two  triangles  and  add  their  areas. 
II.    .'.   construct 

Let  the  pupil  complete  the  analysis. 

Proof: 

STATEMENTS 

1.  Area  ABC  =  i  ab. 

2.  Area  ADC  =  ia'b\ 

3.  Area  ABC+avea,  ADC  =  i  ab+i  a'b'. 

4.  a  =  a'. 

5.  Area  ABCD  =  \  a{b+b'). 

Let  the  pupil  give  the  reasons. 

Ex.  1.     One  base  of  a  trapezoid  is  3  ft.  more  than  the  other. 
If  its  altitude  is  6  ft.  and  its  area  81  sq.  ft.,  find  the  bases. 

Ex.  2.     The  area  of  a  trapezoid  is  96  sq.  ft.,  its  altitude  8  ft., 
one  base  9  ft.     Find  the  other  base. 

Ex.  3.     The  area  of  a  trapezoid  is  the  product  of  the  altitude 
and  the  median  drawn  between  the  non-parallel  sides. 

"  Ex.  4.     There  are  other  methods  of  drawing         i) Q 

construction  lines  in  the  trapezoid  ABCD  (Fig.        /j  "\ 
388)  so  as  to  divide  it  into  parts  whose  areas     j.  f        e 
may  be  found  and  added.     Prove  Th.  115  by  Fig.  38S 

means  of  Fig.  388  and  suggest  other  possible  figures. 


AREA  AND   EQUIVALENCE 


219 


MEASUREMENT   OF   IRREGULAR   POLYGONS 

246.  The  finding  of  the  area  of  a  field  shaped  like  an 
irregular  polygon  is  one  of  the  important  problems  that  a 
surveyor  must  solve.  One  method  frequently  used  is  to 
divide  the  field  into  triangles  and  apply  the  method  for 
finding  the  areas  of  the  triangles. 


Ex.  1.     Compute  the  area  of  the  field  shown 
in  Fig.  389  from  the  following  data: 

AC  =  270  It.    ;/  =  104ft.     k  =  82it. 

Ex.  2.  Compute  the  area  of  the  field  shown 
in  Fig.  390  from  the  following  data: 

BD=U  ft.  6  in.  ^£=16  ft.  3  in.  PM  = 
13  ft.  CI  =  9  ft.  4  in.  BL=  11  ft.  2  in. 
EK=\0  ft.  6  in. 


Sometimes  the  field  to  be  surveyed  is  bounded  on  one  side 
by  a  stream,  the  shore  of  a  lake,  or  a  curved  road.  In  such 
cases  a  straight  line  is  run  near  the  curved  boundary,  and 
the  inclosed  figure  is  cut  into  trapezoidal  shaped  figures  by 
offsets  run  perpendicular  to  the  line,  as  shown  in  Fig.  391. 

In  such  cases  care  should  be  taken  that  the  curved 
boundaries  of  the  figures  are  as  nearly  straight  as  possible. 

Ex.  3.    Find  the  area  inclosed  between  the  fence  AF  (Fig.  391) 
and  the  river  from  the  following  data: 
yl 5  =  150  ft.     DE  =  250  ft.     CC  =  180  ft. 
5C=140ft.     £F=100ft.    DD'  =  95it. 
CD=160ft.    BB'=UO(t.    ££'  =  120  ft.  Pig.  391 

Ex.  4.  The  distances  along  the  straight  line  (AF,  Fig.  391)  are 
usually  made  equal,  as  the  computation  is  then  much  easier. 
Compute  the  area  referred  to  in  Ex.  3  if  the  offsets  are  run  as 
indicated  below.     Draw  the  figure  to  scale. 

Distances  on  ylF        0      100      200      300      400      500      600 
Length  of  offsets      55       76        83        80        50        42        65 


220  PLANE  GEOMETRY 

Ex.  5.    Find    the    area    of    the    field  shown    in    Fig.    392. 
^5  =  300  ft.,  BC  =  UO  ft.,  ZB  is  a  right  ,^ 

angle,  DE=  125  ft.     Offsets  are  run  every  yV  \\ 

50  ft.  from  D  to  A.    ^D  =  310  ft.     The  y^  j\c 

lengths  of  the  offsets  are  ^^-'^^^^^^'^^^^ 

at  D  .    .     0     at  150  ft.   .    .    .  20  Z^^!^"'"''"^ 

at  50  ft.     .    .  22     at  200  ft.  .    .    .     0  ^^^=^^— '^ 

at  100  ft.   .    .  26     at  250  ft.   ...   18  Fig.  392 


Ex.  6.     Find    the    area  shown    in    Fig.      I  \  -^ 

393,   using  the  dimensions  given.     A  BCD  is      I    \  7 

a  rectangle.     AE  =  BF.  i — \    *^'  /- 

Fig.  393 

Ex.  7.     Two  streets  intersect  at  right  angles.    A     f1||||pi||r" 
third  street  cuts  the  other  two  at  angles  of  30°  and  60°.         LiF 
The  shortest  side  of  the  triangular  park  left  is  200  feet.       ^B'^ 
If  the  streets  are  60  feet  wide  each,  find  the  area  of      W 
pavement  at  their  intersection.  / 1 

Fig.  394 

The  area  of  an  irregular  polygon  may  be  found  approxi- 
mately by  weighing.  Cut  the  figure  and  also  a  square  unit 
from  the  same  sheet  of  paper  or  cardboard  and  weigh  them 
both.  The  areas  have  the  same  ratio  as  the  weights.  Archi- 
medes used  this  method  to  find  the  areas  of  certain  figures. 
Surveyors  sometimes  use  it  to-day. 

EQUIVALENT  POLYGONS 
TESTS  FOR  EQUIVALENCE 

246.  The  following  notations  will  be  used: 

For  A  ABC  or  EJABCD,  S  for  area,  b  for  base,  and  a  for 
altitude. 

For  AA'B'C  or  EJA'B'CD',  S'  for  area,  h'  for  base,  and 
a'  for  altitude. 

A  preliminary  test  for  equivalent  polygons  was  given  in 
§238.  Others  are  given  in  the  assumptions  of  §237.  The 
following  tests  are  now  evident  and  are  here  stated  formally 
for  convenience : 


AREA   AND  EQUIVALENCE  221 

Test  a:  for  equivalent  triangles  or  parallelograms.  Th. 
116  is  really  a  corollary  of  Ths.  113  and  114. 

Theorem  116.  Two  parallelograms  or  two  triangles  are 
equivalent  if 

1.  They  have  equal  bases  and  are  between  the  same 

parallels. 

2.  a  =  a^  and  b  =  b\ 

3.  ab  =  a'h'. 

Test  b:  for  triangles  and  parallelograms.  Th.  117  is  a 
corollary  of  Ths.  113  and  114. 

Theorem  117.  If  a  triangle  and  a  parallelogram  have 
equal  bases  and  equal  altitudes,  the  triangle  is  equivalent 
to  half  the  parallelogram. 

By  the  assumptions  of  §237  any  two  polygons  are 
equivalent  if  they  are  sums,  dijBferences,  or  equal  parts  of 
equivalent  polygons. 

EXERCISES  IN  TRANSFORMATION 
247.     1.  Find  in  Fig.  395  four  equivalent  parallelograms. 

2.  How    many    equivalent    parallelo-   >/      jkh  g  f     ed         c 
grams  can  be  constructed  with  their  bases    \     ,'  \    \^ 
on  two  given  parallels?  \\/-^\',C^ 

3.  Using  any  OJABCD,  construct  two 

Fir    'iQ'i 
parallelograms  equivalent  to  it,  (1)  using 

AB  as  the  common  base,  and  (2)  using  AD  a.s  the  common  base. 

4.  What  equivalent  parallelograms  can 
be  found  in  Fig.  396?  Segments  with  the 
same  letters  are  equal.  The  lines  are  par- 
allel as  indicated:  k\\h,  m  \\  n,  r  \\  s. 

5.  Transform  EJABCD  into  a  paral- 
lelogram having  its  base  equal  to  AB  and 
one  side  equal  to  a  given  segment.  Is  the 
problem  always  possible?  Fig.  396 

6.  Transform  OJABCD  into  a  parallelogram  having  its  base 
equal  to  AB  and  one  angle  equal  to  a  given  angle.  Is  this  problem 
always  possible? 


222 


PLANE  GEOMETRY 


Fig.  397 


7.  How  many  equivalent  triangles  can  you  find  in  Fig.  397? 
The     segments    lettered     alike     are 
equal.     The  lines  are  parallel  as  in- 
dicated:  k  II  h  and  w  ||  ». 

Transform  a  given  ^ABC  into 

8.  A  triangle  having  the  base 
equal  to  AB  and  one  angle  equal 
to  a  given  angle. 

9.  A  triangle  having  the  base  equal  to  BC  and  one  side  equal 
to  a  given  segment. 

10.  An  isosceles  or  a  right  triangle  having  the  base  equal  to 
(1)  AB,  (2)  BC,  (3)  AC. 

11.  A  triangle  having  two  sides  equal  to  given  segments. 
Suggestion.     Two  transformations  are  necessary. 

248.  The  transformations  in  the  previous  section  may  be 
performed  by  means  of  an  algebraic  analysis. 

Problem  18.     To  transform  a  given  parallelogram  into 
a  rectangle  which  shall  have  a  given  segment  as  its  base. 


Fig.  398 
Given  the  EJ  ABCD  with  h  its  base  and  a  its  altitude,  and 
b'  the  given  segment.     Let  P  represent  the  CO  ABCD. 
To  transform  CJ  ABCD  into  a  rectangle  with  h'  as  its  base. 
Analysis  and  construction: 
Let  R  represent  the  rectangle,  a'  its  altitude. 
I.  To  construct  R  =  P,  construct  a'  so  that  ah  =  a'h'. 

IL    .*.  construct  a'  so  that  -r  —  ~/' 

h      a 

in.    .'.  construct  a  fourth  proportional  to  6',  6,  and  a. 

Let  the  pupil  make  the  construction  in  full  and  give  proof. 


AREA  AND   EQUIVALENCE  223 

Problem  19.  To  transform  a  given  parallelogram  into 
a  square. 

Analysis  and  construction: 

I.  To  construct  the  square  =  P,  construct  x  so  that  ab  =  x^, 
where  x  is  the  unknown  side  of  the  square. 

II.    .'.construct  x  sl  mean  proportional  between  a  and  b. 

Ex.  1.  Transform  a  given  triangle  into  (1)  a  rectangle  having 
a  given  base,  (2)  a  square. 

Ex.  2.  Transform  a  given  parallelogram  into  an  isosceles 
triangle  having  a  given  base. 

Ex.  3.  Construct  a  square  which  shall  be  equivalent  to  three 
times  a  given  square. 

Ex.  4.  Construct  on  a  given  base  an  isosceles  triangle  which 
shall  be  equivalent  to  twice  a  given  square. 

249.  Problem  20.  To  transform  a  given  quadrilateral 
ABCD  into  a  triangle  with  the  base  in  the  line  AB  and  the 
vertex  at  point  D. 


Fig.  399 

Given  the  quadrilateral  ABCD. 

To  transform  it  into  a  triangle  with  its  base  in  the  line 
AB  and  its  vertex  at  D. 
Analysis  and  construction: 

I.  Since  the  base  is  to  be  in  the  line  AB  and  the  vertex 
at  D,  AABD  will  be  a  part  of  the  required  triangle. 
.*.  transform  ADBC  into  a  triangle  with  DB  for  one 
side  and  another  side  on  AB  extended. 
II.   .'.  construct  a  line  from  .C  \\  BD  and  join  DE. 

Let  the  pupil  name  the  required  triangle  and  give  analysis  for  the 
proof  and  then  the  proof. 


224  PLANE   GEOMETRY 

Ex.  1.     Transform  a  given  quadrilateral  A  BCD  into  a  triangle 
that  shall  have 

a.  Its  base  in  the  line  AB  and  vertex  at  C. 

b.  Its  base  in  the  line  BC  and  vertex  at  A . 

c.  Its  base  in  the  line  BC  and  vertex  at  D. 

d.  Its  base  in  the  line  AD  and  vertex  at  C. 

e.  Its  base  in  the  line  DC  and  vertex  at  B. 

Problem  21.     To    transform    a    given    polygon    into    a 
triangle. 


A  B 

Fig.  400 

The  analysis,  directions,  and  proof  are  left  to  the  pupil.  The  figure 
suggests  the  construction.  Let  the  pupil  extend  the  method  so  as  to 
transform  a  given  hexagon  into  a  triangle. 

Ex.  2.     Construct  a  square  equivalent  to  a  given  4-side. 

Ex.  3.  Construct  on  a  given  base  a  rectangle  equivalent  to  a 
given  4-side. 

Ex.  4.  Construct  on  a  given  base  an  isosceles  triangle  equiva- 
lent to  a  given  4-side. 

EXERCISES  INVOLVING  EQUIVALENT  FIGURES 

250.  1.  The  median  of  a  triangle  divides  it  into  two  equiva- 
lent parts. 

2.  Divide  a  given  triangle  into  three  equivalent  parts  by  seg- 
ments drawn  from  the  vertex  to  the  base. 

3.  The  diagonals  of  a  parallelogram  divide  it  into  four  equiva- 
lent triangles. 

4.  Given  the  CJABCD  with  X  any  point 
in  the  diagonal  AC.  Prove  that  AAXD  = 
AAXB  (see  Fig.  401). 

5.  Investigate  the  case,  Ex.  4,  in  which ^'^ 
point  X  is  on  ^C  extended.  Fig.  401 


AREA  AND  EQUIVALENCE  225 

6.  Given  A  BCD,  any  quadrilateral,  with  E  and  F  the  mid- 
points  oi    AB   and    CD   respectively.      Prove 
AECF=AAFD-^AEBC  (Fig.  402). 

Suggestion.     Draw  AC. 

7.  Two  triangles  are  equivalent  if  two  sides 
of  one  are  equal  respectively  to  two  sides  of  the 
other  and  the  angles  included  by  these  sides  are  supplementary. 


8.  In  Fig.  403,  A  BCD  is  a  trapezoid  with  its 
diagonals  AC  and  DB.    Prove  AAOD  =  ABOC. 
Suggestion.    First  compare  AABD  and  A  ABC  or  a.^ 


AADCand.  ABDC.  Fig.  403 

9.  In    Fig.    404,    AE\\DB.     Prove   AABE  = 
AAED;    ABDE  =  ABDA;    AABO=AEDO\ 
AADC  =  BCDE.  . , 

Fig.  404 

10.  Show  how  to  divide  a  triangle  into  four  equivalent  parts. 
Can  this  be  done  in  more  than  one  way? 

11.  If  one  base  of  a  trapezoid  is  twice  the  other,  the  diagonal 
divides  the  trapezoid  into  two  parts  one  of  which  is  double  the 
other. 

12.  A  BCD  is  a  parallelogram.  E  and  F  are  the  mid-points 
o(  AB  and  AD  respectively.  Prove  that  CF,  CA,  and  CE  divide 
the  parallelogram  into  4  equivalent  parts. 

13.  Show  how  to  divide  a  parallelogram  into  six  equivalent 
parts  by  lines  drawn  from  the  same  vertex. 

14.  Show  how  to  divide  a  parallelogram  into  three  equivalent 
parts  by  lines  drawn  from  the  same  vertex. 

15.  An  isosceles  right  triangle  is  equivalent  to  K  of  the  square 
constructed  on  its  hypotenuse. 

16.  The  figure  formed  by  joining  the , mid-points  of  the  sides  of 
a  square  is  equivalent  to  J4  of  the  square. 

Suggestion.    Construct  the  medians  of  the  given  square. 

17.  Is  Ex.  16  true  for  any  parallelogram? 


226  PLANE  GEOMETRY 

THE  SQUARE  ON  THE  HYPOTENUSE  OF  A 
RIGHT  TRIANGLE 

261.  Theorem  118.  The  square  constructed  on  the 
hjrpotenuse  of  a  right  triangle  is  equivalent  to  the  sum  of 
the  squares  constructed  on  the  other  two  sides. 


E      M  D 

Fig.  405 

Hypothesis:  ABC  is  a  rt.  A  with  ZA  =  1  rt.  Z  and  the 
m  I,  II,  and  III  constructed  on  the  sides  AB,  AC^  and  BC 
respectively. 

Conclusion:  nUl=\JI+ nil. 

Analysis  and  construction: 

I.  Dili  can  be  proved  equivalent  to  DI+DII  by 
dividing  Dili  into  two  parts  equivalent  to  DI 
and  nil  respectively. 
II.  One  way  of  dividing  D  III  into  the  two  parts  required 
is  by  drawing  a  line  from  A  ±  BC  and  proving 
BEML  =  D I  and  CDML  =  D II. 
III.  To  prove  CDML  =  D II,  compare  them  with  con- 
gruent triangles. 

ly.  The  triangles  desired  may  be  obtained  by  joining 
AD  and  BK.     Then  prove 

1.    AACDmBCK.        2.  AACD  =  HCDML. 
3.    ABCK  =  }4  OIL 


AREA  AND  EQUIVALENCE 


227 


V.  To  prove  AACD  ^  ABCK,  prove  AC  =  CK,  BC  =  CD, 
ZACD=ZKCB, 

Vr.  To  prove  AACD  =  H  CDML,  show  that  CD  is  the 
common  base  and  CL  is  equal  to  the  altitude  of 
each. 
VII.  To  prove  ABCK  =  M  □  II,  show  that  CK  is  the  com- 
mon base  and  AC  is  equal  to  the  altitude  of  each. 
VIII.  To  prove  AC  equal  to  the  altitude  of  ABCK,  prove 
that  HAB  is  a  straight  Hne. 

Let  the  pupil  give  the  analysis  to  prove  that  BEAfL^DI,  full 
details  of  the  proof,  and  the  conclusion. 

For  other  methods  of  proof  see  §255;  §254,  Ex.  23;  §233, 
Ex.  23. 

Problem  22.  To  construct  a  square  equivalent  to  the 
sum  of  two  given  squares. 


Fig.  406 
Given  D I  and  D II  constructed  on  the  segments  a  and  b 
respectively. 

To  construct  a  square  equivalent  to  ni+  nil. 
The  solution  is  left  to  the  pupil. 

Problem  23.     To  construct  a  square  equivalent  to  the 
difference  between  two  given  squares. 
The  solution  of  this  problem  is  left  to  the  pupil. 
Construct  a  square  equivalent  to 
Ex.  1.    The  sum  of  three  given  squares. 
Ex.  2.    The  sum  of  two  given  parallelograms. 
Ex.  3.     The  diderence  between  two  given  triangles. 
Ex.  4.    The  sum  of  a  given  triangle  and  a  given  rectangle. 
Ex.  5.     If  a  and  b  are  two  given  segments,  construct  x  so  that 
;c=  Va'^-l-62;  so  that  x=  yla^-b'^. 


228  PLANE  GEOMETRY 

SUMMARY  AND   SUPPLEMENTARY  EXERCISES 
252.  SUMMARY    OF    IMPORTANT    POINTS    IN    CHAPTER    X 

A.  Formulae  obtained. 

I.  Area  of  rectangle  =  a6  (§239). 
II.  Area  of  parallelogram  =  a&  (§  242) . 

III.  Area  of  triangle  =  3^  a6  (§243). 

IV.  Area  of  trapezoid  =  M  a  (b+b')  (§  244). 

V.  If  a  and  b  are  the  legs  and  c  the  hypotenuse  of  a  right 

triangle,  a'^-^b^  =  c''  (§251). 
For  area  of  triangle  see  §253,  Ex.  41  and  Ex.  44. 
Ex.41.     S  =  }i  be  sin  A. 
Ex.  44.     S=  ^s{s  —  a)  (s—b)  (s  —  c)  where  a,  b,  and  c  are 

sides  of  A. 
For  areas  of  irregular  polygons  see  §245. 

263.     EXERCISES  INVOLVING  NUMERICAL  COMPUTATIONS 

Note.     Be  prepared  to  prove  the  theorems  on  which  any  of   the 
following  exercises  depend. 

1.  A  rectangle  whose  base  is  81  ft.  has  the  same  area  as  a  square 
whose  side  is  36  ft.     Find  the  difference  between  the  perimeters. 

2.  The  altitudes  of  two  triangles  are  equal  and  their  bases  are 
25  ft.  and  40  ft.  respectively.  What  is  the  base  of  a  triangle 
equivalent  to  their  sum  having  an  altitude  2}^  times  as  great? 

3.  The  base  and  altitude  of  a  triangle  are  18  ft.  and  24  ft. 
respectively.  At  a  distance  of  10  ft.  from  the  base  a  line  is  drawn 
parallel  to  the  base.  Find  the  area  of  the  two  parts  into  which  the 
triangle  is  divided. 

4.  Find  the  altitude  of  a  triangle  with  base  21  in.  which  has 
the  same  area  as  a  parallelogram  whose  base  is  18  in.  and  whose 
altitude  is  15  in. 

5.  The  area  of  a  rhombus  is  ^  the  product  of  the  diagonals. 

6.  Find  the  area  of  a  rhombus  if  the  sum  of  the  diagonals  is 
12  ft.  and  their  ratio  is  3  :  5.     Use  Ex.  5. 

'  7.  A  rectangular  field  is  30X80  ft.  It  is  surrounded  by  a  road 
■of  uniform  width  the  entire  area  of  which  equals  the  area  of  the 
field.     Find  the  width  of  the  road. 


AREA  AND  EQUIVALENCE 


229 


8.  The  legs  of  a  right  triangle  are  15  ft.  and  20  ft.  A  perpen- 
dicular is  drawn  from  the  vertex  of  the  right  angle  to  the  hypotenuse. 
Find  the  areas  of  the  two  triangles  formed. 

9.  The  area  of  a  rectangle  is  120  sq.  ft.;  one  side  is  8  ft.  Find 
the  diagonal. 

10.  The  perimeter  of  a  rhombus  is  100  ft.;  the  shorter  diagonal 
is  14  ft.    Find  the  area.     Use  Ex.  5. 

11.  The  area  of  a  rhombus  is  2184  sq.  ft.;  the  shorter  diagonal 
is  26  ft.     Find  the  longer  diagonal  and  one  side.     Use  Ex.  5. 

12.  A  house  is  45  ft.  long,  32  ft.  wide,  24  ft.  to  the  roof,  and 
30  ft.  to  the  ridgepole.  Find  the  number  of  sq.  ft.  in  the  entire 
exterior  surface. 

13.  Find  the  area  of  a  right  triangle  if  its  perimeter  is  84  ft. 
and  its  sides  are  in  the  ratio  3:4:5. 

14.  The  height  of  a  lean-to  roof 

is  8  ft.;    the  span  is  15  ft.     Find 

the  area  of  the  roof  if  the  length  is  ^ 

21  ft.  (see  Fig.  407). 

Fig.  407 

15.  The  parallel  sides  of  an  isosceles  trapezoid  are  30  ft.  and 
40  ft.  respectively;  the  non-parallel  sides  are  each  13  ft.  Find 
the  altitude  and  the  area. 

JO 

16.  The  area  of  a  kite  is  K  the  product  of  the 
diagonals. 

17.  In  Fig.  408,  /LA  =  l  rt.  Z.  ^5  =  ^Z)and 
^C  is  the  perpendicular  bisector  of  DB.  If  ^5  =  4 
in.  and  AC  =  ^}i  in.,  find  the  area  of  A  BCD 
and  the  length  of  DC. 

18.  The  design  shown  in  Fig.  409  is  symmet- 
ric with  regard  to  both  diagonals  of  the  square. 
If  AY  =  )4  AB  and  OX  =  }iOA,  find  the  area  of 
the.  kite  AYXZ  and  hence  of  the  Maltese  cross 
shown,  if  ^5  =  6  in. 

19.  Upon  the  diagonal  of  a  rectangle  11  ft.  wide  and  60 
ft.  long  a  triangle  is  constructed  whose  area  equals  the  area  of  the 
rectangle.     Find  its  altitude. 


A      Y 

Fig. 409 


16 


230 


PLANE   GEOMETRY 


D     IT 

T     C 

V 

m^ 

s 

\V 

^W^ 

^w 

z 

A 

y  3 

20.  One  side  of  a  right  triangle  is  8  in.  and  the  hypotenuse  is 
17  in.  Find  the  area  of  an  equilateral  triangle  constructed  on  the, 
third  side. 

21.  The  diagonals  of  two  squares  are  10  ft.  and  15  ft.  respec- 
tively.  Find  the  diagonal  of  a  square  equivalent  to  their  sum. 

22.  Fig.  410  represents  a  square  with  each  side  ^'r-"""-'^-^^ 

divided  into  8  equal  parts.     What  fraction  of  the 

given  square  is  each  of  the  irregular  figures  shown? 

If  yl 5  is  6  in.,  find  the  length  of  the  sides  of  each 

figure.  ^ 

^  Fig.  410 

Note:  Modern  standard  tiles  are  made  in  certain  definite  sizes 
and  shapes,  all  of  which  are*  derived  directly  from  the  six-inch  square. 
Fig.  410  shows  three  such  tiles. 

23.  Fig.  411  is  a  square  with  the  lines  drawn 
symmetric  with  respect  to  both  medians.  If 
AB=\2  in.  and  AX  =  ?>  in.,  find  the  area  of  the 
shaded  portion. 

24.  Find  the  area  of  an  equilateral  triangle  if 
each  side  is  7  in.;  9  in.;  12  in.;  if  each  side  is  s. 

25.  If  the  area  of  an  equilateral  triangle  is  60,  find  one  side. 

26.  Find  the  area  of  a  regular  hexagon  if  each  side  is  4  in. 

27.  The  sides  of  a  regular  hexagon  are  extended  until  they 
intersect.  Find  the  area  of  the  star  formed  if  each  side  of  the 
hexagon  is  3  cm. 

28.  Upon  the  altitude  of  an  equilateral  triangle  as  a  side  a  second 
equilateral  triangle  is  constructed.  Find  the  area  of  the  second 
triangle  if  a  side  of  the  first  is  6  in. 

29.  Find  the  area  of  an  isosceles  right  triangle  if  the  hypotenuse 
is  6  in.;  4  in.;  8  in.;  10  in.;  if  the  hypotenuse  is  a. 

30.  In  Fig.  412,  ABC  is  an  isosceles  right 
triangle.  ED  is  parallel  to  BC.  F  is  the 
mid-point  of  ED.  FG  and  FH  are  parallel  to 
AB  and  AC  respectively.  If  C5  =  9  in.  and 
^Z)  =  4>^  in.,  find  the  areas  of  the  various  parts 
into  which  the  figure  is  divided. 


Fig.  411 

From  a  parquet 

floor  design. 


AREA  AND  EQUIVALENCE 


231 


Fig.  414 


31.  Fig.  413  represents  a  square  with  each  side  divided  into  4 
equal  parts  and  the  points  joined  as  indicated.  If  AB  =  12  in.» 
find  the  area  of'  the  star. 

i>( g- T lO 

32.  Draw  a  figure  similar  to  Fig.  413,  dividing      L-fe^<^-J 
each  side  into  3  equal  parts.     Find  the  area  of  the 

star  formed  if  each  side  of  the  square  is  6  in.  p-^-^^^3^^. 

33.  A  square  and  an  equilateral  triangle  have  each    -*^    ^ " — ^ 

a  perimeter  of  60  ft.    Find  the  ratio  of  their  areas.  ^^^'  ^^^ 

34.  Fig.  414  shows  a  square  with  each  side  divided 
into  three  equal  parts  and  the  points  joined  as  indi- 
cated. If  each  side  of  the  original  square  is  12  in., 
find  the  area  of  the  eight-pointed  star  and  of  the 
irregular  octagon  in  the  center. 

Note.  The  designs  shown  in  Figs.  413  and  414  are  extensively  used 
in  industrial  ornament.  Fig.  415  is  from  a  parquet  floor  design.  See 
Fig.  181a.  -Dciw ^ ^(? 

35.  Fig.  415  shows  a  square  with  AX  and  CY 
each  one-fourth  the  diagonal  AC.  If  the  points  are 
joined  as  indicated,  prove  that  XBYD  is  a  rhombus 
and  find  its  area.    AB  =  Q  in.  Fig.  415 

36.  If  the  parallel  sides  of  a  trapezoid  are  8  in.  and  20  in.  and 
its  altitude  is  6  in.,  find  the  area  of  the  two  triangles  formed  by 
extending  the  non-parallel  sides  until  they  intersect. 

37.  In   Fig.   410,  XF  =  50,   ZF  =  20,   ZW  =  ob, 

y4X  =  63,    YD  =120,  BZ  =  SO,  aF  =  00.     Find  the 

area  of  A  BCD.     AX,  DY,  BZ,  and  CW  are  each 

perpendicular  to  XW.  x rz — ht 

Fig.  416 

38.  A  BCD  is  a  quadrilateral  with  AC  and  D  rt.  A  and 
ZA  =  }4  Tt.  Z.  If  BC  =  8  in.  and  ^Z>=15  in.,  find  the  area  of 
the  quadrilateral. 

39.  The  area  of  one  rectangle  is  3  times  the  area  of  a  second. 
If  the  base  of  the  first  is  twice  the  base  of  the  second,  find  the 
relation  between  their  altitudes. 

40.  The  area  of  a  certain  rectangle  is  3K  times  the  area  of 
a  triangle.  If  the  base  of  the  triangle  is  twice  the  base  o^  the 
rectangle,  find  the  relation  between  their  altitudes. 


232  PLANE   GEOMETRY 

41.  lie  and  b  represent  two  sides  of  a  triangle  and  A  the  included 
angle,  prove  that  the  altitude  upon  side  ft  is  c  sin  ^ ,  that  the  altitude 
upon  side  cis  b  sin  A,  and  that  the  area  is  >2  be  sin  A. 

Find  the  area  of  AABC,  using  the  trigonometric  tables,   if 

42.  .4j5  =  8.3cm.,    ^C  =  3.9cm.,     ZA=37°. 

43.  ^5  =  9. 6cm.,     ^C  =  5. 4cm.,     ZA=54°, 
t44.  If  a,  b,  and  c  are  the  sides  of  a  triangle, 

the  area  of  the  triangle  is  -^sis  —  a)  {s  —  b)  (s  —  c),  ^^ 
where  s  =  }4{a+b-\-c).  Fig.  417 

A  nalysis: 
I.  To  find  the  area  of  a  triangle  in  terms  of  the  three  sides,  find 
the  area  in  terms  of  the  base  and  altitude  and  express  the 
altitude  in  terms  of  the  three  sides  by  means  of  the  Pythago- 
rean theorem. 

II.  Ave3.  =  y2ch  (Fig.  417). 

III.  h^  =  b^-x\ 

IV.  h^  =  a^-{c-xy. 

Solve  the  two  equations  for  h  and  x  and  substitute  the  value  of  h  in 
the  area  formula. 

Outline  of  algebraic  proof: 

i2-x2  =  a2_(c-x)2 

b^-x^  =  a'^-c^+2cx-x'^ 


\       2c 

(2bc-b^-c^+a^)    (2bc+b^+c^-a^) 
4c2 

,,,     (a-b+c)   (a+b-c)   (b+c+a)   (b+c-a) 

"  ~  4c^ 

This  may  be  put  in  a  more  concise  form  by  noting  that 

a+b—c  =  a-\-b+c—2c 

a-b+c  =  a+b+c-2b 

b-\-c—a  =  a-\-b-{-c—2a 
We  may  let  a-\-b-^c=2s  for  convenience. 
Then  a-\-b-c=2s-2c  =  2{s-c) 
a-b+c  =  2s-2b  =  2{s-b) 
b-{-c-a  =  2s-2a  =  2{s-a) 


AREA  AND  EQUIVALENCE  233 


Substituting  in  the  value  for  /t', 

^,     2(s-b)  .  2{s-c)  .  25  .  2{s-a) 

^  4c^  

,     2  V5(5-a)  (s-b)  (s-c) 
n  = 


/.  area  of  A  =  5  ch=  "^ s{s—a)  (s  —  b)  {s—c),  where  s  =  ^{a-^b-{-c). 

Note.  What  changes  must  be  made  in  the  details  above  if  h  falls 
outside  the  triangle? 

Note.  This  formula  was  given  by  a  noted  Greek  engineer  and 
surveyor,  Hero  of  Alexandria  (about  100  B.C.). 

45.  Find  the  areas  of  triangles  whose  sides  are : 

a.  13,  14,  15  c.  15,  18,  21 

b.  8,  10,  12  d.  24,  33,  41 

46.  Compute  from  the  following  data  the  area  of  the  field 
shown  in  Fig.  418,  using  the  formula  given  in  Ex.  44.  ^ 

A  B  =  210  ft.  DE  =  305  it. 
BC  =  210  it.  EA  =225  it. 
CD  =  210  it.        BD  =  SMit. 

AD  =  ^25  it. 

* 

EXERCISES  INVOLVING  EQUIVALENT  POLYGONS 

254.  Note.  Be  prepared  to  prove  the  theorems  on  which  any  of 
the  following  exercises  depend. 

1.  Make  a  review  diagram  for  Th.  117. 

2.  The  area  of  a  triangle  is  one-half  the  product  of  the  perimeter 
and  the  radius  of  the  inscribed  circle. 

3.  The  area  of  any  polygon  circumscribed  about  a  circle  is  one- 
half  the  product  of  the  perimeter  and  the  radius 
of  the  inscribed  circle. 

4.  Given  the  isosceles  right  A  ABC  with  OF  and 
GE  drawn  from  the  mid-point  oi  AC  parallel  to  AB 
and  BC  respectively,  prove  area  EBFG=}/2  area 
ABC  (see  Fig.  419).  "     F  J.  419 

5.  The  line  joining  the  mid-points  of  the  parallel  sides  of 
a  trapezoid  divides  it  into  two  parts  that  have  equal  areas. 


234 


PLANE   GEOMETRY 


6.  Given  A^  BC  with  A"  an  arbitrary  point  on  the 
median  CO.     Prove  AACX  equivalent  to  ABCX. 

7.  The  three  medians  of  a  triangle  divide  it  into  6 
equivalent  triangles. 

8.  Given  AADC  and  DEE  with  AD  =  DB,  DE  = 
EC,  ZADC+ZBDE  =  2  rt.A,  prove  area  DBE  =  y> 
area  ADC  (Fig.  420). 

9.  Given  A  ABC  with  CD,  an  arbitrary  segment 
from  C  to  AB,  divided  into  three  equal  parts  at  points 
X  and  Y,  compare  area  AYB,  area  AXB,  and  area 
ACB  (Fig.  421). 

10.  Construct  a  square  whose  area  shall  be  H  of 
the  area  of  a  given  square. 

11.  In.  Fig.  422,  ABC  is  an  isosceles  right  tri- 
angle. AE  =  EB;  BF  =  FC;  £G  and  Ffi^  are  per- 
pendicular to  AC.  Compare  areas  AEG,  EBF, 
EFHG,  and  ABC. 


Fig. 


421 


*    12.  In  Fig.  423,  ABC  is  an  isosceles  A.     AC 

and  CB  are  each  divided  into  3  equal  parts.     CD,     

FM,  and  GN  are   A_  AB.     Compare  the  areas  of  ^      fJg.^423      ^ 
the  various  figures  formed.  King-rod  and 

Queen-rod  truss 
design 

13.  Given  the   quadrilateral   A  BCD   with    the     ^ 
diagonal  BD  bisected  at  X,  prove  that  AX  CD  is 
equivalent  to  AXCB  (see  Fig.  424). 

14.  Transform  quadrilateral  A  BCD  into  quad- 
rilateral ABCE  so  that  (1)  Z  ECB  shall  be  equal  to 
a  given  angle,  (2)  side  AE  shall  be  equal  to  a  given 
segment.     Is  the  problem  always  possible? 


Fig.  424 


15.  A  quadrilateral  is  bisected  by  one  of  its  diago- 
nals if  the  second  diagonal  is  bisected  by  the  first. 


16.  Given  AABC  with  EF\\AB. 
equivalent  to  ABCE  (Fig.  425). 


Prove  AACF 


Fig.  425 


AREA  AND  EQUIVALENCE 


235 


17.  In  Fig.  426,  A  BCD  is  a  square  with  its 
diagonals  and  medians.  The  semi-medians,  OE, 
OF,  OG,  and  OH,  are  bisected  at  X,  F,  Z,  and  W, 


respectively, 
the  square. 


Prove  that  the  star  formed  is 


18.  Construct  in  a  given  square  a  star 
that  shall  be  ^i  of  the  given  square. 

19.  If  through  any  point  on  the  diagonal 
of  a  parallelogram  segments  are  drawn  par- 
allel to  the  sides,  the  parallelograms  on  oppo- 
site sides  of  the  diagonal  are  equivalent  (see  Fig.  427). 

20.  In  Fig.   428,  ABC   is   any  triangle,     [sj 
AC  ED  and   BFEC  have   the   common   side  CE 
and  have  AC  and  BC  as  bases.     Prove  that  if  d 
Z?F  is  joined,  ABFD  is  a  parallelogram  equivalent 
to  the  sum  of  [sJ  AC  ED  and  BFEC.  p^^   ^28 

Suggestion.     Extend  EC  and  show  that  CO  ABFD 
can  be  broken  up  into  parts  equivalent  to  EJACED  and  CO  BFEC. 

21.  In  Fig.  429,  ABC  is  any  triangle.  ACFG  ^ 
and  BCED  are  any  parallelograms  on  the  sides 
AC  and  BC  respectively.  GF  and  DE  are 
extended  to  meet  at  M.  On  AB  a  parallelogram 
is  constructed  with  one  side  AH  equal  and  paral- 
lel to  MC.  Prove  that  CJABKH  is  equivalent 
to  CJ ACFG-]- CJ BCED. 

Note.     Ex.  21  is  known  as  Pappus'  theorem  (about  300  a.d.). 

22.  Show  hoUr  to  construct  a  parallelogram  equivalent  to  the 
sum  of  two  given  parallelograms. 

23.  Deduce  the  Pythagorean  theorem  from  Pappus'  theorem. 

D_  0  c 

24.  In  Fig.  430,  A  BCD  is  a  square  with  its  diago- 
nals and  medians.  HE,  EF,  FG,  and  GH  are  each 
divided  into  3  equal  parts  and  the  points  are  joined 
as   indicated.     Compare   the   area   of   the   shaded 

portion  with  the  area  of  the  square  A  BCD.  a        e^ 

Fig.  430 


: JSL. : 


236 


PLANE  GEOMETRY 


25.  In  Fig.  431,   ABCD  is  a  square  with  its 
diagonals  and  medians.     OX,  OY,  OZ,  and  OW 
are   each    H    of   the  corresponding  semi-diagonal,  ^'f^^f^^ 
Compare  the  area  of  the  star  with  the  area  of  the 

square  ABCD.  a         ji         b 

Fig.  431 
Note.     Figs.  426,  430,  and  431  are  from  parquet  floor  designs. 

26.  If  the  mid-points  of  two  adjacent  sides  of  a  parallelogram 
are  joined,  the  triangle  formed  is  equivalent  to  M  the  parallelo- 
gram. 

27.  If  the  mid-points  of  two  sides  of  a  triangle  are  joined  to 
any  point  of  the  base,  the  quadrilateral  formed  is  equivalent  to 
3^  the  triangle. 

28.  The  triangle  formed  by  joining  the  mid-point  of  one  of 
the  non-parallel  sides  of  a  trapezoid  with  the  extremities  of  the 
opposite  side  is  equivalent  to  >2  the  trapezoid. 

29.  Given  the  EJ ABCD  and  0,  any  point  within  the  parallelo- 
gram, prove  that  /\AOB  -\-  ADOC  is  equivalent  to  K  EJABCD. 

30.  If  in  Ex.  29  point  0  were  without  the  OJ ABCD,  what 
would  be  the  relation  between  AAOB  and  DOC  and  EJ ABCD} 

31.  In  Fig.  432,  AABC  is  isosceles.  CD  is 
perpendicular  to  AB.  AB,  AC,  and  CB  are 
each  divided  into  3  equal  parts  and  the  points 
are  joined  as  indicated.  Compare  the  areas  of 
the  triangles  formed. 

t32.  Construct  a  rectangle  so  that  its  area 
shall  be  equal  to  the  area  of  a  given  square  ^" 

and  the  sum  of  its  base  and  altitude  equal  to  ;      I   a^  I 

a  given  segment  (see  Fig.  433).  I ' 

A  nalysis: 

Let  X  represent  the  base  and  y  the  altitude  of  the  rectangle. 

To  construct  a  rectangle  so  that  its  area  is  equal  to  the  area  of  the 

X     a 
square,  construct  x  and  y  so  that  xy  =  a^,  or  so  that  -  =  -  • 

/,  x-\-y  must  be  divided  into  two  segments  so  that  a  is  a  mean  pro- 
portional between  x  and  y. 

Directions  and  proof  are  left  to  the  pupil. 


F 

Fig.  432 

From  a  roof  truss 

design 


Fig.  433 


AREA  AND  EQUIVALENCE 


237 


t33.  Construct  a  rectangle  so  that  its  area  shall  be  equal  to 
the  area  of  a  given  square  and  the  difference  between 
its    base    and    altitude    equal    to    a    given    segment 
(see  Fig.  434). 

Suggestion.  Let  a  be  one  side  of  the  square  and  x 
and  y  the  base  and  altitude  of  the  rectangle.  An  analysis 
similar  to  that  for  Ex.  32  will  show  that  a  will  be  the 
mean  proportional  between  two  segments  x  and  y  whose 
difference  is  -45  (see  Fig.  434). 


134.  If  two  triangles  have  an  angle  of  one  equal 
to  an  angle  of  the  other,  the  ratio  of  the  areas 
equals  the  ratio  of  the  products  of  the  sides  that 
include  the  equal  angles  (see  Fig.  435). 


Fig.  434 


Fig.  435 


area  ABC 


be 


.  „,^,  =  Tj-y '  compare  the  area  of  each  triangle  wi 
area  A B'C      be  ^ 


th 


A  nalysis: 
I.  To  prove 

the  area  of  a  third  triangle  having  the  same  altitude. 

TT      .   J         /-D/      J  c   J  4.U        ^-      .area  ABC         ,     .   area  AB'C 

II.    . .  draw  LB  and  find  the  ratio  of .  „,>;  and  of t-^ttt^, 

area  A  B'C  area  A  B'C 

and  multiply. 
Let  the  pupil  give  the  proof. 

35.  If  s  and  s'  represent  the  areas  of  the  A  ABC  and  A  B'C 
shown  in  Fig.  435,  find 

1.  c',iib  =24,  b'  =  15,c=lS,  and  s  =  s'. 

2.  c',  if  b  =56,  6'  =  35,  c  =  70,  and  s  =  2s'. 

3.  b,  if  6'=  18,  c'  =  28,  c  =  45,  and  s=r2s'. 

36.  Given  a  triangular  field  A  BC,  show  how  to  divide       ^/jy 
it  into  two  equivalent  parts  by  running  a  fence  from 
point  X  on  one  side  (see  Fig.  436). 

Fig.  436 

Suggestion.     Let    Y  be  the  point  at  which  the  fence  meets  BC. 
Represent  ^  C  by  b,  XC  by  b\  CF  by  a',  and  5C  by  a.     Point  Y  and 


.'.  a'  are  to  be  found, 
struction  in  full. 


Find  a'  so  that 


ab_ 

a'b'     1 


Why?     Give  con- 


238  PLANE  GEOMETRY 

EXERCISES  INVOLVING  THE  PYTHAGOREAN  THEOREM 


256.  1.  In  the  details  of  the  proof  of 
Th.  lis  as  given  above,  prove  that 
ACBF  and  ABE  have  equal  bases  and 
equal  altitudes  instead  of  proving  them 
congruent  (Fig.  437). 


Fig.  437 


A  nalysis: 

I.  To  prove  that  ACBF  and  AABE  have  equal  bases  and  equal 
altitudes,  use  BF  and  AB  as  bases  and  prove  the  altitudes 
equal. 

drop  perpendiculars  from  E  and  C  to  ^B  and  BF  extended 
respectively  and  prove  ABEP^  ABRC. 


II. 


2.  Give  the  details  for  the  proof  for  Th.  118  (1)  if  square  CE  is 
constructed  on  the  opposite  side  of  CB  from  that  shown  in  Fig.  437; 
(2)  if  square  AF  is  constructed  on  the  opposite  side  oi  AB  from 
that  shown  in  the  figure;  (3)  if  both  CE  and  AF  are  on  opposite 
sides  of  CB  and  AB  respectively  from  thovse  shown  in  the  figure. 

3.  Show  how  Fig.  438  may  be  used  to  prove 
the  Pythagorean  theorem.  A  EH  is  the  given 
right  triangle.  Extend  AE  and  AH,  making  AB  = 
AD  =  a-\-b,  Complete  the  square  on  a+b.  Make 
BF  =  CG  =  DH  =  a.  Join  E,  F,  G,  and  H.  From  E 
and  F  draw  lines  parallel  to  AD  and  AB  respec- 
tively.    Prove  that  EFGH  is  a  square  equivalent  to  a^-\-b^ 


438 


4.  Show  how  Fig.  439  may  be  used  to  prove 
the  Pythagorean  theorem.  ABE  is  the  given  right  x*,^ 
triangle.  Construct  the  square  on  AB.  Make 
AK  =  BE.  JoinDK.  Construct  A DCF  and  A C5G 
congruent  to  AABE.  Prove  KHFD  and  BGHE 
squares  whose  sum  is  equivalent  to  c^.  Fig.  439 

5.  Show  how   Fig.  439   may  be  constructed  by  constructing 
squares  BGHE  and  KHFD  first. 

6.  If  a  and  b  are  two  given  segments,  show  by  geometry  that 
'4a^  +  b'^  is  not  a+b. 


AREA  AND  EQUIVALENCE  239 

7.  If  a,  b,  and  c  are  three  given  segments,  construct  a  fourth 
segment  x  so  that  (I)  x=  ^|a^^¥Tc^,  (2)  x=  yja^  +  f^-c^. 

8.  Construct  a  square  whose  area  shall  be  H  of  the  difference 
between  the  areas  of  two  given  squares. 

9.  Construct  on  a  given  base  a  right  triangle  whose  area  shall 
be  equal  to  the  sum  of  the  areas  of  two  given  triangles. 

10.  AB  is  the  diameter  of  a  semicircle.  D  is  a  point  on  ^5  so 
located  that  AD=/i  AB.  DE  is  ±  AB  a.t  D  and  cuts  the  semi- 
circle at  E.    Prove  that  the  square  on  ^£  is  H  the  square  on  AB. 

1 1 1 .  If  ^  BC  is  any  triangle  with  Z  A  an  S. 

acute  angle,  and  sides  a,  6,  and  c  are  the'  \.y^  'V 

sides  opposite   AA,  B^  and  C  respectively,  y^         l''   \ 

a'^  =  ¥-\-c'^  —  2cp  where  p  is  the  projection  of  ^^^ — ^.L—. — I A^ 

6  upon  c  (Fig.  440).  Fig.  440 
Outline  of  algebraic  proof:     Let  h  be  the  altitude  on  c. 
a^  =  h^-{-(c-p)^  ^ 

Note.  The  distance  between  the  feet  of  the  perpendiculars  drawn 
to  a  given  line  from  the  ends  of  a  segment  is  called  the  projection  of 
the  segment  upon  the  line. 

12.  Give  the  proof  for  Ex.  11  for  a  figure  in 
which  h  falls  on  AB  extended.  ^^, 

tl3.  If  ABC  is  any  triangle  with  Z  A  any  >''^^/i^   ''^ 

obtuse  angle,  and  sides  a,  b,  and  c  are  opposite     ^^      /        I 

A  A,   B,  and   C  respectively,   a^  =  b'^-\-c^-{-2cp  b        c :/—-p—J 

where  p  is  the  projection  of  b  upon  c  (Fig.  441).  ^^^^-  ^^^ 

Outline  of  algebraic  proof:    Let  h  be  the  altitude  upon  c. 

a2  =  A2  +  (c+p)2 

h2  =  b^-p2 

...  a^  =  i,i^p2^(^c-{-pr- 

14.  State  and  prove  the  converse  of  the  proposition  that  the 
square  on  the  hypotenuse  of  a  right  triangle  equals  the  sum  of 
the  squares  of  the  other  two  sides. 

15.  If  the  sides  of  a  triangle  are  as  given  below,  is  the  largest 
angle  right,  acute,  or  obtuse:  (1)  20,39,36?  (2)  15,30,39?  (3) 
15,  36,  39  ? 


CHAPTER  XI 
Similarity 

INTRODUCTORY 

DEFINITION 

256.  Similar  polygons  have  been  defined  as  polygons  that 
have 

1.  The  angles  of  one  equal  to  the  corresponding  angles  of 
the  other,  and 

2.  Corresponding  sides  proportional. 

By  the  ratio  of  similitude  of  two  similar  polygons  is  meant 
the  ratio  of  any  two  corresponding  sides.  For  convenience 
corresponding  sides  will  be  lettered  alike,  as  AB  and  A'B\ 
BC  and  B'C. 

TESTS  FOR   SIMILAR   POLYGONS 
TEST  I  FOR   SIMILAR  TRIANGLES 

257.  The  first  test  for  similar  triangles  is  given  in  Th.  102. 
Two  mutually  equiangular  triangles  are  similar. 

Let  the  pupil  review  the  proof  for  Th.  102. 

Ex.  1.  Two  triangles  are  similar  if  two  angles  of  one  are 
equal  respectively  to  two  angles  of  the  other. 

Ex.  2.  Two  right  triangles  are  similar  if  an  acute  angle  of 
one  is  equal  to  an  acute  angle  of  the  other. 

Ex.  3.  Two  isosceles  triangles  are  similar  if  an  angle  of  one  is 
equal  to  a  corresponding  angle  of  the  other. 

Ex.  4.  Construct  two  similar  triangles  with  the  ratio  of 
similitude  %. 

Ex.  5.  Two  triangles  similar  to  a  third  are  similar  to  each 
other. 

240 


SIMILARITY 


241 


TEST  n  FOR   SIMILAR  TRIANGLES 
258.    Ex.  1.    Given  ABC  any  triangle,  and  A'B'  of  AA'B'C, 


construct  AA'B'C  so  that  Z^  =  Z^'  and 


AB      AC 


A'B'    A'C 

Theorem  119.  Two  triangles  are  similar  if  an  angle  of 
one  is  equal  to  an  angle  of  the  other  and  the  ratios  of  the 
including  sides  are  equal. 


Hypothesis: 

A'B'    A'C' 
Conclusion: 


Fig.  442 

In   AABC  and   AA'B'C,    ZA=  ZA'  and 


AABCooAA'B'C. 


Analysis  and  construction: 
L  To  prove  AABC co  AA'B'C ,  prove   AB  =  ZB'  and 

/:c=zc'. 

II.  To  prove  ZB=  ZB',  place  AABC  upon  AA'B'C 
so  that  the  sides  of  Z  A  are  collinear  with  the  sides 
of  ZA'  and  prove  BC\\B'C'. 

III.  To  prove  BC\\  B'C,  prove  ^,  =  ^,. 

The  proof  is  left  to  the  pupil. 

Ex.  2.  Construct  AA'B'C  similar  to  AABC  if  the  sum  of 
sides  A'B'  and  B'C  is  given. 

Ex.  3.  If  a  segment  is  drawn  parallel  to  the  base  of  a  triangle 
and  terminated  by  the  sides,  the  triangle  formed  is  similar  to  the 
given  triangle.    Give  two  proofs. 

Ex.  4.  Solve  Ex.  3  if  the  parallel  to  the  base  cuts  the  sides  of 
the  triangle  extended. 


242  PLANE  GEOMETRY 

TEST  III  FOR   SIMILAR   TRIANGLES 

259.    Ex.  1.     Given  A  ABC  any  triangle.     Construct  AA'B'C 
A  7?        J^r*        A  (^ 
SO  that  jr^'^'^X'^AX^'  ^^^^^  '^'^'  °^  A^'j5'C'. 

Theorem  120.     Two  triangles  are  similar  if  the  corre- 
sponding sides  have  equal  ratios. 


Fig.  443 

Hypothesis:     In  AABC  and  A  A  'B'C\  -  =  -=--> 

cab 

Conclusion:     AABC  ~  A  A  'B'C 
Analysis  and  construction: 

I.  To  prove  AABC<^  AA'B'C ,  compare  each  with  a 

third  triangle. 
II.  On  CA  construct  CX  =  b\  on  CB  construct  CY  =  a\ 
and  prove  7i  ~  Tg  and  Tz  ^  Tg  where  T^  =  AABC, 
T2  =  AA'B'C\  and  T^  ^AXYC. 

III.  To  prove  T2  ^Tg,  prove  c'  =  z.     (Use  Th.  4.) 

IV.  To  prove  c'  =  z,  compare  two  proportions  containing 

z  and  c'\  t-  =  -»  from  A  Ti  and  7„,  and  7-  =  -  given. 
be  ^  be 

b'     z 
V.  To  prove  ^  =->  prove  T^ 00  Tg.     (Use  Th.  119.) 

Let  the  pupil  give  the  proof.     For  IV  use  Th.  92. 

Ex.  2.  Construct  a  triangle  similar  to  a  given  triangle  with 
its  perimeter  equal  to  a  given  segment. 

Ex.  3.  Prove  that  two  triangles  are  similar  if  two  sides  and 
the  median  to  one  of  them  are  proportional  to  the  corresponding 
parts  of  the  other. 


SIMILARITY  243 

CONSTRUCTION  OF  SIMILAR  TRIANGLES 

260.  Problem  24.  Upon  a  given  segment  corresponding 
to  a  side  of  a  given  triangle,  to  construct  a  triangle  similar 
to  the  given  triangle. 

GENERAL   TEST   FOR   SIMILAR   POLYGONS 

261.  Theorem  121.  Two  polygons  are  similar  if  diago- 
nals drawn  from  two  corresponding  vertices  divide  the 
polygons  into  the  same  number  of  triangles  similar  each 
to  each  and  similarly  placed. 


E^A—P 


Fig.  444 

Hypothesis:  Polygons  P  and  P'  are  divided  into  triangles 
by  the  diagonals  drawn  from  the  corresponding  vertices 
A  and  A'  so  that  AI~  AI',  AII~  AlF,  etc. 

Conclusion:     Polygon  P  co  polygon  P'. 

Analysis: 

I   I.  ZB=  IB\  /.C=  AC\  etc. 

To  prove  Pc^^P/,  prove  <  ^^  <^  _^_^ 

(  'a'~~h'~7' 

To  prove  r>  =  -/.  prove  each  ratio  equal  to  777^,- 
be  yi  C 

Let  the  pupil  complete  the  analysis  and  give  the  proof. 

CONSTRUCTION   OF   SIMILAR  POLYGONS 

262.  Problem  25.  Upon  a  given  segment  corresponding 
to  a  given  side  of  a  given  polygon,  to  construct  a  polygon 
similar  to  the  given  polygon. 

Can  this  problem  be  solved  in  more  than  one  way.? 


244 


PLANE   GEOMETRY 


PROPERTIES   OF   SIMILAR  POLYGONS 
COMPOSITION   OF   SIMILAR  POLYGONS 

263.  Theorem  122.  If  two  polygons  are  similar,  diago- 
nals drawn  from  two  corresponding  vertices  divide  the 
polygon  into  the  same  number  of  triangles  similar  each  to 
each  and  similarly  placed. 

The  analysis  and  the  proof  are  left  to  the  pupil. 

RATIOS   OF   CORRESPONDING   SEGMENTS 

264.  Theorem  123.  The  ratio  of  corresponding  altitudes 
of  two  similar  triangles  equals  the  ratio  of  the  bases. 


Fig.  445 


Hypothesis:     In   AABC^  AA'B'C,  h  and  h'  are  corre- 
sponding altitudes  and  c  and  c'  are  the  bases. 


Conclusion :     ri^-r 
h      c 


Analysis:     To  prove  ti  =  ~ 
third  ratio  — ;  • 


h  c 

compare  -t-,  and  —  with   the 


Let  the  pupil  complete  the  analysis  and  give  the  proof. 

Ex.  1.     Prove  that  in  two  similar  triangles  the  ratio  of  the 
following  segments  equals  the  ratio  of  similitude  of  the  triangles: 
a.  Bisectors  of  corresponding  angles. 
h.  Corresponding  medians. 

c.  The  radii  of  the  circumscribed  circles. 

d.  The  radii  of  the  inscribed  circles. 

Ex.  2.  Are  there  any  other  segments  in  two  similar  triangles 
whose  ratio  is  equal  to  the  ratio  of  similitude  of  the  triangles? 
Give  proofs. 


SIMILARITY  245 

Ex.  3.     In  two  similar  polygons  the  ratio  of  corresponding 
diagonals  is  equal  to  the  ratio  of  similitude  of  the  polygons. 

Ex.  4.     If  b  and  b'  are  the  bases  of  two  triangles  and  a  and  a' 

the  corresponding  altitudes,  show  that  -,  =  rr  if  the  triangles  are 

similar  and  — -  =  -—  if  they  are  equivalent. 
a      0 

265.  Theorem  124.  In  a  series  of  equal  ratios  the  sum 
of  the  antecedents  is  to  the  sum  of  the  consequents  as  any 
antecedent  is  to  its  consequent. 

TT        ,     .      a      c      e 
Hypothesis:  T"  =  "7  =  "7  =  etc. 

Conclusion: =  — . 

•      •  6-f-fi+/+ctc.      b 

Proof: 

STATEMENTS 

a      c      a  __  e 

2.  ab  =  ab 
be  =ad 
be  =  af 


3.  ab+bc-{-be-\-etc.  =ab-{-ad-{-af -^etc. 

4.  b{a-{-c+e+etc.)  =a{b+d-{-f +etc.). 
a  +  c  +  e  4-etc.  _ _a  ^ 

^'  b  +^H-/+etc.~  b  ' 

Let  the  pupil  give  all  reasons. 

h       AD        ^  h       DB 
Ex.  1.     Prove  Th.  123  by  provmg  that  Jji^'^r^t  and  ^/  =  ^/^/' 

Theorem  125.     The  ratio  of  the  perimeters  of  two  similar 
triangles  is  equal  to  the  ratio  of  similitude. 

Let  the  pupil  give  the  analysis  and  the  proof.     Use  Th.  124.     The 
series  of  equal  ratios  is  obtained  from  the  ratios  of  corresponding  sides. 

Ex.  2.     Corresponding  diagonals  of  two  similar  polygons  have 
the  same  ratio  as  the  perimeters. 

17 


240 


PLANE  GEOMETRY 


RATIOS   OF  AREAS 

266.  Theorem  126.  The  areas  of  two  similar  triangles 
have  the  same  ratio  as  the  squares  of  the  bases  or  the  squares 
of  the  altitudes. 


Fig.  446 
Hypothesis:     In   AABC^  l\A'B'C ,  h  and  h'  are  corre- 
sponding bases  and  a  and  a'  the  altitudes. 
Area  ABC  ^^_a^ 

52      ^2 
Analysis:     To  prove   _^^  \'i'Dtnt^TJ'i.^~''i"   prove    each 


Conclusion. 


ArQ2i  A' B'C 

area  ABC 


"   ^^"^"   area  A '^'C     h'^     a'' 
ratio  equal  to  a  third  ratio. 

Proof: 

STATEMENTS 

I.  1.  Kve2iABC=y2ah. 
2.  AreaA'^'C'  =  3^a'6'. 
area  ABC         ah 


X      0. 

••areaA'^'C'     a'h'' 

II.  1. 

a     b 
a'     b'' 

2. 

a      a      b     a 

a      b      b      b 
''''  a''  b'~b'  '  b' 

thatis,^,  =  -,^,  =  p-.. 

III.    .-. 

area  ABC      6^      a^ 
areaA'Br'"6'2     a"^ 

Let  the  pupil  give  all  reasons. 

SIMILARITY 


247 


Ex.  1 .  In  two  similar  triangles  let  s  and  5'  represent  the  areas, 
b  and  b'  the  bases,  and  a  and  a'  the  altitudes.     Find 

a.  s',  if  5  =45  and  a=  ^  a'. 

b.  a\  if  a  =16  and  s  =  Hs'' 

c.  b',  if  ^>=  10  and  5' =  35. 

Ex.  2.  The  sum  of  the  areas  of  two  similar  triangles  is  78 
sq.  ft.  Two  corresponding  sides  are  10  ft.  and  15  ft.  Find  the 
areas  and  the  altitudes. 

267.  Theorem  127.  The  areas  of  two  similar  polygons 
have  the  same  ratio  as  the  squares  of  two  corresponding 
sides. 


Fig.  447 

Hypothesis:     In  polygons  P  and  P',  a  and  o'  are  corre- 
sponding sides. 

Area  P  _a^ 


Conclusion  : 


Area  P' 

Analysis  and  construction: 

area  P      a 
I.  To  prove  — '■ — tt,  =  -?.>,  break  the  polygon  up  into  tri- 
^         area  P     a^ 

angles  that  are  similar  in  pairs  and  find  the  ratio  of 

the  sums  of  the  areas  of  the  triangles;  that  is,  draw 

diagonals  from  corresponding  vertices  and  prove 

A-fB+C+ete.     a2 


A'+B'+C'+etc.     a'2 

11.   To  prove  ^,_^^,^^,^^^^  = 

a2                   A        B 
=  ^„  prove  ^,=  ^, 

C 

-etc.,  that  is,  prove     ., 

a^      B      a2 

a'2'    B'     a'2'  '^^" 

and 

then  use  Th.  124. 

Let  the  pupil  give  the  pi  oof. 

248  PLANE   GEOMETRY 

Ex.  1.  If  the  area  of  one  polygon  is  twice  that  of  a  similar 
polygon,  what  is  the  ratio  of  the  bases? 

Ex.  2.  The  area  of  one  polygon  is  three  times  the  area  of 
a  similar  polygon.  If  the  base  of  one  is  15,  find  the  base  of  the 
other. 

Ex.  3.  Two  corresponding  sides  of  two  similar  polygons  are 
5  and  9  respectively.  If  the  area  of  the  first  is  30,  find  the  area 
of  the  second. 

Ex.  4.  The  sum  of  the  areas  of  two  similar  polygons  is  910. 
The  ratio  of  similitude  is  %.     Find  the  area. 

SUMMARY  AND   SUPPLEMENTARY  EXERCISES 
268.  SUMMARY  OF  IMPORTANT  POINTS  IN  CHAPTER  XI 

A.  Tests  for  similarity. 

I.  To  prove  two  triangles  similar,  prove  that 
a.  They  are  mutually  equiangular  (§257). 
h.  Their  corresponding  sides  have  equal  ratios 

(§259). 
c.  An  angle  of  one  equals  an  angle  of  the  other, 
and  the  ratios  of,  etc.  (§258).' 
II.  To  prove  two  polygons  similar,  prove  that  diagonals 
drawn  from  corresponding  vertices,  etc.  (§261). 

B.  Properties  of  Similar  polygons. 

I.  Angles   are  equal    and    corresponding    sides   have 
equal  ratios   (§256). 

IL  Two  similar  polygons  may  be  divided  into  triangles, 
similar  in  pairs  by,  etc.  (§263). 

III.  Ratios  of  corresponding  segments  in  similar  poly- 
gons equal  the  ratio  of  similitude  of  the  polygons 
(§§264,265). 

IV.  Ratio  of  the  areas  of  similar  polygons  equals  the 
square  of  the  ratio  of  similitude  of  the  polygons 
(§§266,  267). 


SIMILARITY  249 

EXERCISES  INVOLVING  SIMILAR  POLYGONS 
269.  1.  Make  review  diagrams  for  Ths.  122  and  127. 
If  the  segments  which  join  the  vertices  of  a  polygon  with  a  given 
point  are  divided  in  the  same  ratio  from  the  given  point  and  the  points 
of  division  joined  in  the  same  order  as  the  vertices  of  the  polygon, 
the  polygon  so  formed  and  the  given  polygon  are  radially  placed. 
The  point  may  be  without  the  polygon  as  in  Fig.  448a,  or  within  the 

polygon  as  in  Fig.  4486.    In  each  case  7yTi  =  7JW}^^^^-  "^^^  radial  point 
is  called  the  center  of  similitude  of  the  polygons. 

D 


o<- 


FiG.  4486 

2.  Two  polygons  that  are  radially  placed  are  similar.  Give 
the  proof  for  the  two  cases. 

3.  Show  how  a  polygon  may  be  constructed  similar  to  a  given 
polygon  A  BCD  with  a  given  segment  A'B'  corresponding  to  side 
AB  and  with  any  arbitrary  point  O  as  a  center  of  similitude. 

Note.     The  center  of  similitude  of  two  similar  figures  may  be  of 
use  in  enlarging  or  reducing  drawings. 

4.  Make  a  drawing  and  give  a  proof  for  Ex.  3  with  point  O 
falling  (1)  on  point  A;    (2)  on  AD. 

5.  If  the  perimeter  of  an  equilateral  triangle  is  66,  find  the 
sides  of  a  similar  triangle  with  half  the  altitude. 

6.  On  one  side  and  on  the  altitude  of  an  equilateral  triangle 
as  bases  construct  similar  triangles.  What  is  the  ratio  of  their 
areas? 

7.  On  one  side  and  on  the  diagonal  of  a  square  construct 
equilateral  triangles.     What  is  the  ratio  of  their  areas? 

8.  Construct  a  triangle  similar  to  a  given  triangle  whose  area, 
shall  be  ^4  the  area  of  a  given  triangle;  ^2  the  area;  14  the  area; 
}i  the  area. 


250  PLANE  GEOMETRY 

t9.  If  the  hypotenuse  of  a  right  triangle  is  twice  the  shorter 
side,  one  of  the  acute  angles  of  the  triangle  is  60°. 

10.  The  triangle  A  BC  has  a  right  angle  at  C,  and  D  trisects 
AC  so  that  AD  =  2DC.  It  is  then  found  that  AD  =  BD.  Find 
the  size  of  angle  A. — College  Entrance  Examination  Board,  Plane 
Geometry  Examination,  1907. 

fll.  If  two  triangles  have  their  corresponding  sides  parallel 
each  to  each,  the  triangles  are  similar. 

tl2.  If  two  triangles  have  their  corresponding  sides  perpen- 
dicular each  to  each,  the  triangles  are  similar. 

13.  Fig.  449  shows  two  quadrilaterals  with  their 
sides  respectively  parallel.  If  AC  \\  A'C,  prove  the 
quadrilaterals  similar. 

14.  Construct  a  quadrilateral  similar  to  a  given  ^^ 
quadrilateral  having  one  diagonal  equal  to  a  given 
segment.  Fig.  449 

15.  The  perimeters  of  two  similar  triangles  are  21  and  c 
The  altitude  of  the  first  is  8.     Find  the  altitude  of  the  second. 

16.  In  Fig.  450,  ABC  is  an  isosceles  triangle. 
CD  J_  from  C  to  the  base  AB.  PQ  is  the  per- 
pendicular bisector  of  AC.  Prove  that  A  A  DC 
is  similar  to  AAPQ  and  find  the  length  of  ^Q 
in  terms  of  ^Z>  and  CD.  "* 

17.  In  Fig.  451,  ABCD  is  a  square.  AE  =  BF= 
CG  =  DH  and  the  points  are  joined  as  indicated. 
Find  all  pairs  of  congruent  triangles  and  of  similar ''^ 
triangles  formed.  Read  the  ratios  between  corre- 
sponding sides  of  the  similar  triangles.  If  AB  =  S6 
cm.  and  AE=y3AB,  find  AX,  XY,  and  YF. 

18.  The  equal  sides  of  an  isosceles  triangle  are  13,  the  base  is 
10.     Find  the  sides  of  a  similar  triangle  whose  altitude  is  4. 

19.  What  is  the  area  in  acres  of  a  portion  of  a  map  that  covers 
4^2  sq.  in.?    The  scale  of  the  map  is  1  inch  =  l  mile. 

20.  The  sides  of  a  polygon  are  10,  15,  9,  and  22.  Find  the 
sides  of  a  similar  polygon  if  its  perimeter  is  140. 


SIMILARITY  251 

t21.  If  similar   polygons   are   constructed   on   the   sides  of  a 
right  triangle  as  corresponding  sides,  the  area  of  the 
polygon  constructed  on  the  hypotenuse  is  equal  to 
the  sum  of  the  areas  of  the  polygons  on  the  other 
two  sides  (see  Fig.  452). 

Analysis:  Fig.  452 

To  prove  that  C=i4-fB,  find^  and  -^  in  terms  of  the  sides  of  the 

triangle,  and  add. 

t22.  To  construct  a  polygon  equivalent  to  one  of  two  given 
polygons  and  similar  to  the  other. 


Fig.  453 
Analysis: 
I.  Suppose  it  is  required  to  construct  a  polygon  C  equivalent  to  B 

and  similar  to  i4. 

A     a* 
II.  Since  Cis  to  be  ^^1  —  =— . 

A     A 

III.  Since  C  is  to  be  equivalent  to  5,  7;  =  ^  . 

IV.  As  A  and  B  are  not  similar,  we  cannot  compare  their  areas  by 

comparing  squares  of  corresponding  sides.     .'.  reduce  A  and  B 
to  equivalent  squares. 

Let  A  be  equivalent  to  a  square  whose  side  is  m. 
Let  B  be  equivalent  to  a  square  whose  side  is  n, 

__    _,,        A     m» 
V.  Then^=-. 

VL    .•.^=^. 
a      n 

VII.    .*.  to  find  a',  find  a  fourth  proportional  to  m,  n,  and  a. 

Let  the  pupil  give  directions  in  full,  make  the  drawing,  and  give 
the  proof. 


CHAPTER   XII 

Regular  Polygons 

DEFINITION 

270.  A  regular  polygon  has  been  defined  as  a  polygon 
with  all  of  its  sides  and  all  of  its  angles  equal. 

Ex.  1.     A  regular  quadrilateral  is  a  square. 

Ex.  2.     An  equilateral  triangle  is  a  regular  polygon. 

CONSTRUCTION   OF  REGULAR  POLYGONS 
GENERAL   THEOREMS 

271.  Theorem  128.  If  a  circle  is  divided  into  n  equal 
arcs,  the  chords  joining  the  points  of  division  form  a  regular 
polygon. 

The  analysis  and  proof  are  left  to  the  pupil. 

272.  Theorem  129.  If  a  circle  is  divided  into  n  equal 
arcs,  the  tangents  drawn  to  the  points  of  division  form  a 
regular  polygon. 


Fig.  434 

Hypothesis:  OO  is  divided  into  the  equal  arcs  WX,  XY, 
YZ,  etc.,  and  tangents  AB,  BC,  CD,  etc.,  are  drawn  to  the 
points  of  division,  X,  Y,  Z,  etc. 

Conclusion:     A  BCD  etc.  is  a  regular  polygon. 

252 


REGULAR  POLYGONS 


253 


Analysis  and  construction: 
I.  To  prove  that  ABCD  etc.  is  a  regular  polygon,  prove 
AA=  Z.B=  ZC;  =  ctc.,  and  >lB  =  5C:  =  CD  =  etc. 
II.  To    prove    that     Z A  =  ZB=  ZC  =  etc.,    and    that 
AB  =  BC  =  CD  =  etc.,  join  WX,  XY,  YZ,  Z\\  etc.. 
and  prove  AWXA  m  AXYB  ^  AYZC  m,  etc. 

Suggestion.    Notice  that 

AX  =  BY=  CZ  =  etc By  congruent  triangles. 

BX  =  CY  =  ZD  =etc By  congruent  triangles. 

AB  =  BC=  CD  =  etc Equal  segments  added  to  equal  segments. 

The  proof  is  left  to  the  pupil. 

CONSTRUCTION  OF  THE  INSCRIBED  SQUARE  AND 
RELATED  POLYGONS 

273..  Problem  2G.    To  inscribe  a  square  in  a  circle. 


No.  1  No.  2  No.  3 

Fig.  455 
Analysis  and  construction  (Fig.  455,  No.  1) : 
I.  To  inscribe  a  square  in  a  circle,  divide  the  circle  into 

4  eqiial  arcs. 
II.   .'.divide  the  perigon  about  the  center  of  the  circle 
into  4  equal  angles. 

III.    .'.construct 

Let  the  pupil  complete  the  directions  and  give  the  proof. 

274.  Problem  27.     To  inscribe  a  regular  octagon  in  a 
circle  (Fig.  455,  No.  2). 

Ex.  1.  What  other  regular  inscribed  polygons  can  be  obtained 
from  the  inscribed  square? 

Ex.  2.  Show  how  to  construct  regular  circumscribed  polygons 
of  4,  8,  and  16  sides  (Fig.  455,  No.  3). 


254  PLANE  GEOMETRY 

CONSTRUCTION   OF  THE   REGULAR   INSCRIBED 
HEXAGON   AND   RELATED   POLYGONS 

276.  Problem  28.     To  inscribe  a  regular  hexagon  in  a 
circle. 


No.S 

Fig.  456 

Analysis  and  construction  (Fig.  456,  No.  1) : 

I.  To  inscribe  a  regular  hexagon  in  a  circle,  divide  the 
circle  into  6  equal  arcs. 

II.    .'.divide  the  perigon  about  the  center  of  the  circle 
into  6  equal  angles. 

III.  Since  H  of  360  is  60,  construct  6  angles  of  60°  each  at 
the  center  of  the  circle. 

IV.    .*.  construct   ..... 

Let  the  pupil  complete  the  directions  and  give  the  proof. 

Ex.  1.  Prove  that  the  side  of  the  regular  inscribed  hexagon 
is  equal  to  the  radius  of  the  circle. 

Ex.  2.     Inscribe  an  equilateral  triangle  in  a  circle. 

Ex.  3.     Inscribe  a  regular  12-side  in  a  circle. 

Ex.  4.  What  other  regular  inscribed  polygons  may  be  obtained 
from  the  regular  inscribed  hexagon? 

Ex.  5.  Show  how  to  construct  regular  circumscribed  polygons 
of  3,  6,  and  12  sides. 

Note.  Regular  polygons  are  in  very  common  use  for  towers,  spires, 
bay  and  dormer  windows,  hoppers,  nuts,  and  the  like.  They  are 
extensively  used  in  ornament.  Even  the  less  common  forms,  such  as 
polygons  with  7,  9,  11,  or  even  13  sides,  occasionally  occur. 


REGULAR   POLYGONS 


255 


CONSTRUCTION   OF   THE   REGULAR   INSCRIBED 
DECAGON   AND   RELATED   POLYGONS 

276.  The  construction  of  the  regular  inscribed  decagon 
depends  directly  upon  the  following  problem. 

Problem  29.  To  divide  a  given  segment  so  that  the  larger 
portion  is  a  mean  proportional  between  the  whole  segment 
and  the  smaller  portion. 


Fig.  457 
Given  the  segment  a. 

Ci  X 

To  find  a  segment  x  so  that  —  = 

X     a  —  x 

Analysis  and  construction: 
I.  To  find  X  so  that  —  = 


a  —X 


solve  for  x. 


1.  a'^—ax  =  x'^ 


2.  x'^-^ax  =  a'^. 

3.  x^+ax+{\aY  =  a^-^(Aa)\ 

4.  {x+\ay  =  a^+{\a)\ 

Equation  4  suggests  the  Pythagorean  theorem. 
II.    .*.  construct  a  right  triangle  with  a  and  \a  for  perpen- 
dicular sides.     The  hypotenuse  will  be  x-\-\a. 
III.    .'.  X  will  be  the  difference  between  the  hypotenuse 
and  \a. 
Outline  of  proof: 

1.  (^  +  ia)2  =  a2  +  Ga)^ 

2.  x'^  +  ax  +  {\ay  =  a'' -Vilay. 


3.  x^-^-ax 

4.  x'' 


i2_ 


ax  =  a{a  —  x) 


5.  ^  =  ^^- 
x    a  —X 


Let  the  pupil  give  the  reasons. 


256  PLANE  GEOMETRY 

If  a  segment  is  so  divided  that  the  larger  portion  is  a  mean 
proportional  between  the  whole  segment  and  the  smaller 
portion,  the  segment  is  said  to  be  divided  into  extreme 
and  mean  ratio.  Problem  30  may  be  stated :  To  divide  a 
segment  into  extreme  and  mean  ratio. 

Note.  The  division  of  a  segment  into  extreme  and  mean 
ratio  is  often  called  the  Golden  Section.  It  was  known  to 
the  Pythagoreans.  It  is  an  interesting  fact,  often  used  in 
the  theory  of  design,  that  that  division  of  a  rectangle  which 
is  the  most  pleasing  to  the  majority  of  people  is  the  one 
that  most  closely  approximates  the  golden  section.    It  might 


be  used  in  designs  for  doors  or  windows  (Fig.  458) .  ^^°-  ^^^ 

Exercise.     Show  that  if  —  =  —3-  the  ratio  —  is  approximately 

8:13. 

277.  Problem  30.    To  inscribe  a  regular  decagon  in  a 
circle. 


Fig.  459 

Given  circle  0. 

To  construct  a  regular  inscribed  decagon. 

Analysis: 
I.  yo  construct  AB,  a  side  of  the  decagon,  construct 
Z0=   Ho  of  4  rt.  ^  or  Vs  of  2  rt.  A. 
:.  construct  A  ABO  isosceles  so  that  ZA=2Z0. 

II.  The  problem  is  reduced  to  the  construction  of  AAOB, 
The  fact  that  ZA  =  2Z0  suggests  drawing  the 
bisector  of  Z  A.    A  ABC  and  ABO  would  then  be 

similar    and    .*.  -r^  =  -^^  or  AB  =0B  '  BC. 


REGULAR  POLYGONS  257 

Construction: 
1.  Divide  the  radius  of  the  circle,  OB,  into  extreme  and 
mean  ratio  at  C. 

IL  Use  the  larger  segment,  OC,  as  a  side  of  the  decagon. 
in.   .-.  make  AB  =  OC.     Join  OA  and  prove  ZO  =  M  of  a 

Proof: 

STATEMENTS 


makeAB  =  OC. 

straight  angle. 

1.  OB 

s 
OC 

OC 

CB 

2.  OB 

AB 

AB 
BC 

3.  AABC^AABO. 

4.  AABC  is  isosceles. 

5.  .*.  A  A  BO  is  isosceles. 

6.  :.AC  =  AB  =  OC. 

7.  .*.  AOC  is  isosceles. 

8.  Z2=Z0. 

9.  .-.  Z3  =  Z0. 

10.  /.  Zi4=twice  ^O. 

11.  Z.4  +  Z5+ZO  =  5XZO  =  180°. 

12.  .-.  ZG  =  H  180°  =  Mo  360°. 
Let  the  pupil  give  the  reasons. 

Ex.  L  Show  that  the  radius  of  a  circle  may  be  divided  into 
extreme  and  mean  ratio  and  therefore  that  a  side  of 
the  regular  inscribed  decagon  may  be  obtained  as 
follows:  Draw  ^5  and  CD,  two  diameters  perpendic- 
ular to  each  other.  Let  E  be  the  mid-point  of  OA . 
With  E  as  center  and  EC  as  radius  cut  OB  at  F. 
OF  is  a  side  of  the  decagon  (Fig.  460). 

Ex.  2.      Construct    with    ruler    and    compasses      Fig.  460 
angles  of  36°,  18°,  54°,  72°. 

Ex.  3.      What    angles    can    you    construct    with    ruler    and 
compasses  without  the  aid  of  a  protractor? 


258  PLANE  GEOMETRY 

278.  Problem  31.     To  inscribe  a  regular  pentagon  in  a 
circle. 

The  analysis  and  the  proof  are  left  to  the  pupil  (Fig.  461,  No.  1). 

Ex.  1.     What  other  regular  inscribed  polygons  can  be  obtained* 
from  the  regular  inscribed  decagon? 

Ex.  2.     Show  how  to  construct  regular  circumscribed  polygons 
of  5  and  10  sides. 

279.  Problem  32.     To  inscribe  a  regular  pentadecagon 
in  a  circle. 

Suggestion.     The  central  angle  must  be  24° :  60"  -36°  =  24°  (Fig.  461, 

No.  2). 

Exercise.     What  inscribed  and  circumscribed  regular  polygons 
may  be  obtained  from  the  regular  inscribed  pentadecagon?     How? 


CONSTRUCTION  OF  OTHER  REGULAR  POLYGONS 

280.  Note.  From  the  time  of  Euclid  until  1796  it  was  supposed 
that  the  regular  polygons  mentioned  in  §§273-279  were  the  only  ones 
that  could  be  constructed  with  ruler  and  compasses.  This  includes 
polygons  with  3  •  2",  4  •  2",  5  *  2",  and  15  •  2'*  sides.  The  smaller  poly- 
gons not  included  in  this  set  are  those  of  7,  9,  11,  13,  14,  17,  18,  19 
sides.  In  179G,  however,  Karl  Friedrich  Gauss,  then  a  young  man 
of  nineteen  in  one  of  the  German  universities,  proved  that  regular 
polygons  with  a  prime  number  of  sides  could  be  inscribed  in  a  circle 
by  means  of  a  ruler  and  compasses  if  and  only  if  the  prime  number  was 
of  the  form  2"  +  l.  That  no  polygons  of  7,  9,  11,  13,  19,  etc.,  sides 
can  be  constructed  in  the  given  manner  follows  from  Gauss's  proof. 
A  polygon  of  7  sides  can  be  constructed  by  the  use  of  a  parabola  and 
a  circle;  one  of  9  sides  by  the  use  of  a  hyperbola  and  a  circle. 


REGULAR   POLYGONS 


259 


281.  While  regular  polygons  of  7,  9,  11,  13,  etc.,  sides 
cannot  be  constructed  exactly  with  ruler  and  compasses, 
methods  for  constructing  them  approximately  are  frequently 
given  in  courses  in  mechanical  drawing.  The  accuracy  of 
these  methods  can  be  tested  by  trigonometry. 

PROPERTIES  OF  REGULAR  POLYGONS 

THE   CIRCUMSCRIBED    CIRCLE 

282.  Theorem  130.  A  circle  can  be  circumscribed  about 
any  regular  polygon. 


Fig.  462 


A  BCD  etc.  is  any  regular  polygon. 

A  circle  can  be  circumscribed  about  polygon 


Hypothesis: 

Conclusion: 

A  BCD  etc. 

Analysis: 

I.  To  prove  that  a  circle  can  be  circumscribed  about 
ABCD  etc.,  prove  that  a  point  can  be  found  which 
is  equally  distant  from  the  vertices. 
XL  .'.construct  perpendicular  bisectors  of  any  two  con- 
secutive sides,  BC  and  CD.  Let  the  bisectors 
meet  at  O  and  prove  that  OB  =  OC  =  OD  =  OE  =  etc. 

III.  To  prove  that  OS  =  0C,  prove 

IV.  To  prove  that  OC  =  OZ),  prove  ..... 

V.  To  prove  that  OE  =  OB,  prove  ABOC  mADOE. 
Let  the  pupil  complete  the  analysis  and  give  the  proof. 

Cor.     The  radius  of  the  circumscribed  circle  of  a  regular 
polygon  bisects  the  angle  through  whose  vertex  it  passes. 


260 


PLANE  GEOMETRY 


THE   INSCRIBED    CIRCLE 

283.  Theorem  131.     A  circle  can  be  inscribed  in  any 
regular  polygon. 


Fig.  463 

Hypothesis:     ABODE  etc.  is  any  regular  polygon. 
Conclusion:    A     circle    can    be    inscribed    in    polygon 
ABCDE  etc. 

Analysis  and  construction: 
I.  An  inscribed  circle  will  be  tangent  to  the  sides  of  the 
polygon, 
.'.the  perpendiculars  from  the  center  to  the  sides  of 

the  polygon  must  be  radii  and  .*.  equal, 
.'.construct  the  circumscribed  circle,  draw  the  per- 
pendiculars OX,  OY,  OZ,  etc.,   from  the  center 
to  the  sides,  and  prove  OX  =  OY  =  OZ  — etc. 
The  proof  is  left  to  the  pupil. 

Suggestion.     Show  that  Th.  131  may  be  proved  by  Th.  67  or  by 
Th.  130  Cor.  and  Th.  85. 


11. 


III. 


PROPERTIES  DEPENDENT  UPON  THE  CIRCUMSCRIBED  AND 
INSCRIBED  CIRCLES 

284.  The  center  of  the  circumscribed  and  of  the  inscribed 
circle  of  a  regular  polygon  is  called  the  center  of  the  polygon. 

The  radius  of  the  circumscribed  circle  of  a  regular  polygon 
is  called  the  radius  of  the  polygon. 

The  radius  of  the  inscribed  circle  of  a  regular  polygon  is 
called  the  apothem  of  the  polygon. 

By  the  central  angle  of  a  regular  polygon  is  ineant  the 
angle  between  two  consecutive  radii. 


REGULAR  POLYGONS  261 

Cor.  I.  The  central  angle  of  a  regular  polygon  of  n  sides 
is  Vn  of  360°. 

Cor.  n.  The  radius  of  a  regular  polygon  bisects  the 
angle  between  two  consecutive  apothems,  and  the  apothem 
bisects  the  angle  between  two  consecutive  radii. 

Cor.  Ill  The  radius  of  a  regular  polygon  bisects  the 
arc  between  the  points  of  contact  of  the  inscribed  circle. 

Ex.  1.     Any  two  diagonals  of  a  regular  pentagon  are  equal. 

Ex.  2.  Two  diagonals  from  the  same  vertex  of  a  regular 
pentagon  trisect  the  angle  at  that  vertex. 

Ex.  3.  Show  how  to  divide  a  regular  hexagon  into  two  con- 
gruent isosceles  trapezoids,  three  congruent  rhombuses,  or  six 
congruent  equilateral  triangles. 

Ex.  4.  On  a  given  base  construct  a  regular  hexagon  without 
constructing  the  circumscribed  circle. 

Ex.  5.  A  principal  diagonal  of  a  regular  hexagon 
passes  through  the  center  of  the  circumscribed  circle 
and  is  parallel  to  a  pair  of  opposite  sides  (Fig.  464). 

Ex.  6.  If  a  regular  polygon  has  an  even  number 
of  sides,  the  diameter  of  the  circumscribed  circle  drawn 
from  any  vertex  passes  through  the  opposite  vertex. 

Ex.  7.  If  a  regular  polygon  has  an  even  number  of  sides,  the 
opposite  sides  are  parallel. 

Ex.  8.  If  any  regular  polygon  has  an  odd  number  of  sides,  the 
diameter  of  the  circumscribed  circle  drawn  from  any  vertex  is  a 
perpendicular  bisector  of  the  opposite  side. 

Ex.  9.  A  side  of  an  inscribed  equilateral  triangle  bisects  the 
radius  drawn  to  the  mid-point  of  the  subtended  arc. 

Ex.  10.  The  central  angle  of  a  regular  polygon  is  the  supple- 
ment of  the  interior  angle  of  the  polygon. 

Ex.  11.  The  area  oC  a  square  circumscribed  about  a  circle  is 
twice  the  area  of  the  square  inscribed  in  the  same  circle. 

Ex.  12.  If  squares  are  described  outwardly  on  the  sides  of  a 
regular  hexagon,  the  outer  vertices  of  the  squares  are  the  vertices 
of  a  regular  duodecagon. 

18 


262  PLANE   GEOMETRY 

THE  ANGLE  OF  A  REGULAR  POLYGON 

285.  Theorem  132.     Each  angle  of  a  regular  polygon  of 

.,       .    271-4    .      . 

n  sides  is rt.  A. 

n 

The  proof  is  left  to  the  pupil. 

THE  AREA  OF  A  REGULAR  POLYGON 

286.  Theorem  133.     The  area  of  a  regular  polygon  is 
one-half  the  product  of  the  perimeter  and  the  apothem. 

Analysis: 

To  find  the  area  of  a  regular  polygon  draw  the  radii  and 
add  the  areas  of  the  triangles  formed. 

SIMILAR  REGULAR  POLYGONS 
TEST  FOR   SIMILAR  REGULAR  POLYGONS 

287.  Theorem  134.     Two  regular  polygons  of  the  same 
number  of  sides  are  similar. 


Hypothesis:     The  two  regular  polygons  O  and  G'  h::vc  the 
same  number  of  sides. 

Conclusion:     Polygon  0~  polygon  0\ 
Analysis: 
I.  To  prove  polygon  0^^ polygon  0',  prove  ZA=  LA', 

,  I.      .  r>/      .  .ABBCCD 

ZB=  ZB'  =  etc.,  and  J7^f  =  ^T^f  =  -^^f  =  ^^^- 

^^    ^  AB      BC  AB     A'B'     ^ 

IL  To  prove -^7^,  =  ^,^'  pro ve -^^  = -^^  =  1 . 

The  proof  is  left  to  the  pupil. 


REGULAR   POLYGONS  263 

RATIO    OF   CORRESPONDING   SEGMENTS 

288.  Theorem  135.  If  two  regular  polygons  have  the 
same  number  of  sides,  the  ratio  of  the  perimeters  is  equal 
to  the  ratio  of  the  radii  or  of  the  apothems. 


Fig.  466 

Hypothesis:  In  the  two  regular  polygons  O  and  O'  with 
the  same  number  of  sides,  r  and  r'  are  the  radii,  a  and  a'  the 
apothems,  p  and  p'  the  perimeters,  and  5  and  s'  are  sides. 

Conclusion:     —,  =  -,'=—,• 
p'     r'     a' 

Outline    of   proof: . 

s  _r 
J'~? 
r__a 
?~a' 

Let  the  pupil  make  an  analysis  and  give  all  the  reasons  in  the  proof. 

Cor.  The  ratio  of  the  perimeter  to  the  diameter  of  the 
inscribed  or  of  the  circumscribed  circle  is  the  same  for  all 
regular  polygons  of  the  same  number  of  sides. 

RATIOS   OF  AREAS 

289.  Theorem  136.  If  two  regular  polygons  have  the 
same  number  of  sides,  the  ratio  of  the  areas  is  equal  to  the 
ratio  of  the  squares  of  the  radii  or  of  the  apothems. 

The  analysis  and  the  proof  are  left  to  the  pupil. 


264  PLANE   GEOMETRY 

SUMMARY  AND   SUPPLEMENTARY  EXERCISES 
290.  SUMMARY   OF   IMPORTANT   POINTS   IN   CHAPTER   XH 

A.  Construction  of  regular  polygons. 

I.  To  construct  a  regular  inscribed  or  circumscribed 
polygon  of  n  sides,  divide  the  circle  into  n  equal 
arcs  by  constructing  an  angle  ^/n  of  360°  at  the 
center  of  the  circle  (§§271  and  272). 
II.  To  construct  regular  4-,  8-,  or  18-sided  polygons, 
construct  two  perpendicular  diameters  (§§273, 
274). 

III.  To  construct  regular  3-,  6-,  or  12-sided  polygons, 

construct  a  central  angle  of  60°  by  means  of  an 
equilateral  triangle  (§  275). 

IV.  To  construct  regular  £-,  10-,  or  15-sided  polygons, 

divide  the  radius  of  the  circle  into  extreme  and 
mean  ratio  (§§277,  278,  279). 

B.  Properties  of  regular  polygons. 

I.  A  circle  can  be  circumscribed  about,  etc.  (§282). 
II.  A  circle  can  be  inscribed  in,  etc.  (§  283). 

III.  The  radius  of  the  circumscribed  circle  of  a  regular 

polygon  bisects,  etc.  (§282). 

2w—  4 

IV.  Each  angle  of  a  regular  polygon  is  rt.   A 

ft 

(§285). 
V.  The  central  angle  of  a  regular  polygon  is  V«  of  360° 
(§284). 

C.  The  area  of  a  regular  polygon  is  ^  per.    X  apothem 

(§286). 

D.  Similar  regular  polygons. 

I.  Regular  polygons   are   similar  if  they  have,   etc. 

(§287). 
II.  The  ratio  of  the  perimeters  equals,  etc.  (§288). 
III.  The  ratio  of  the  areas  equals,  etc.  (§289). 
For  similar  polygons  in  general  see  §268. 


REGULAR  POLYGONS 


265 


Fig.  467 


EXERCISES  INVOLVING   PROPERTIES   OF   AND   SPECIAL 
CONSTRUCTIONS   FOR  REGULAR   POLYGONS 

291.  1.  In  a  regular  hexagon  the  secondary  diagonals  FD  and 
AC  are  parallel  to  each  other  (Fig.  467). 

2.  In  Fig.  467,  prove  that  UVWXYZ  is  a  regu- 
lar hexagon. 

Note.  The  star  shown  in  Fig.  467  is  extremely  com- 
mon. It  seems  to  be  an  ancient  symbol  of  Deity.  It 
is  used  in  such  modern  instances  as  the  policeman's 
star  and  many  trademarks. 

3.  The  area  of  the  inscribed  equilateral  triangle  is  half  the  area 
of  the  regular  hexagon  inscribed  in  the  same  circle. 

4.  Show  that  the  area  of  a  regular  hexagon  inscribed  in  a  circle 
is  a  mean  proportional  between  the  areas  of  the  inscribed  and 
circumscribed  equilateral  triangles. 

5.  The  central  angle  of  a  regular  octagon  is  45°. 


6.  Given  a  side  of  a  regular  octagon,  to  construct 
the  circumscribed  circle. 

Suggestion.     See    Fig.    468.     In    order    to    construct 
Zl=22>^^  construct  Z2=45°. 


Fig.  468 


7.  Given  the   secondary  diagonal   AB  oi  a  regular  octagon 
(Fig.  469),  to  construct  the  octagon. 

Suggestion.  Since  the  number  of  degrees  in  Zl  is 
known,  AAEB  may  be  constructed.  The  problem  is 
then  to  construct  DCllAB  so  that  DC=AD. 


Note.  Ex.  7  is  a  problem  which  might  occur  in 
building  octagonal  bay  windows.  ADCB  would  be  the 
outline  of  the  window  and  AB  the  line  on  the  house.      Fig.  469 


8.  Show  that  a  regular  octagon  may  be  inscribed 
in  a  square  by  the  following  method  (Fig.  470): 
With  O  as  center  and  OF,  the  half  median,  as 
radius  cut  the  diagonals  at  F,  Z,  W,  and  X.  At 
F,  Z,  W,  and  X  construct  perpendiculars  to  the 
diagonals. 


J/ 


-*j>-- 


Fig.  470 


266 


PLANE  GEOMETRY 


9.  Show  that  a  regular  octagon  'may  be  inscribed  in  a  square 

by  the  following   method    (Fig.  471):    With   the     j^    ^ s_o 

vertices  of  the   square  as  centers  and  the  semi-      '^'^^^ 
diagonal  as  radius  cut  the  sides  of  the  square  at 
Xy  Y,  Z,  etc.,  and  join  the  points  as  indicated. 


Pig.  473 

Note.  Exs.  8  and  9  are  extremely  useful.  They  can  be  used 
as  shown  in  Fig.  472  to  construct  a  tiled-floor  pattern  composed  of 
regular  octagons  and  squares,  or  as  in  Fig.  473  to  cut  a  square  timber 
down  into  an  octagonal  one.  The  cuttings  at  the  sawmill  would  be 
along  the  lines  AB  and  BX,  DC  and  CY,  and  so  around  the  timber. 
Ex.  9  gives  the  common  method  for  cutting  out  an  octagonal  table  top. 

10.  Fig.   474   shows   an   inscribed   regular  pentagon  with  its 
diagonals.     Prove 

{\)AX  =  XB  =  BY^Qtc.    {2)AB  =  AY.    {Z)  AC  ^ 

is   divided   into   extreme    and    mean  ratio   at    F. 
(4)  ^  F  is  divided  into  extreme  and  mean  ratio  at  X. 

CB 


Suggestion   for 
CB=AY. 


(3). 


Prove 


that    ^ 


CY 


and 


11.  From  the  foregoing  exercises  show  how  to  construct  the 
diagonal  of  a  regular  pentagon,  given  one  side,  and  therefore  how  to 
construct  the  regular  pentagon,  given  one  side. 

12.  Show  that  if  AD  \^  taken  as  the  radius  of  a  circle,  AB  will 
be  a  side  of  the  regular  inscribed  decagon  (Fig.  474). 

13.  Prove  that  VWXYZ  is  a  regular  pentagon  (Fig.  474). 

14.  The  segments  joining  the  mid-points  of  the  sides  of  a  regular 
pentagon  in  order  form  a  regular  pentagon. 

Note.  The  star  shown  in  Fig.  474  is  called  the  pentagram  star 
and  was  used  as  a  symbol  of  recognition  among  the  Pythagoreans, 
an  ancient  Greek  brotherhood  that  studied  geometry.  They  called  it 
Health.     It  is  also  the  star  used  in  the  flag  of  the  United  States. 


REGULAR  POLYGONS  267 

292.     EXERCISES  INVOLVING  THE  MEASUREMENT  OF 
REGULAR  POLYGONS 

1.  Show  that  if  a  square  is  inscribed  in  a  circle  of  radius  1 
the  side  of  the  square  is  V2,  and  that  if  the  radius  is  R  the  side  of 
the  square  is  i?  V2.     What  is  the  area  of  the  inscribed  square? 

2.  Find  one  side  of  a  regular  octagon  inscribed  in 
a  circle  of  radius  1,  without  the  tables. 

Suggestion.     In  Fig.  475,  ^45  is  a  side  of  the  inscribed 

square  and  ^4  C  of  the_inscribed   regular  octagon.     If 

A0  =  l,AD  =  D0  =  yzyl2.     Find  CD  and  .'.  ^ C. 

Fig.  475 

3.  Find  one  side  of  the  regular  inscribed  octagon  if  the  radius 
of  the  circle  is  R. 

4.  Find  one  side  and  the  apothem  of  an  inscribed  equilateral 
triangle  if  (1)  the  radius  of  the  circle  is  1 ;  (2)  the  radius  is  R. 

5.  Find  one  side  of  a  regular  duodecagon  in- 
scribed in  a  circle  of  radius  1,  without  the  tables. 

Suggestion.  In  Fig.  476,  ^5  is  a  side  of  the  inscribed 
regular  hexagon  and  BC  a.  side  of  the  inscribed  regular 
duodecagon.  If  BO  =  l,  AB  =  1,  and  BD  =  }4,  then 
0D  =  }4^^'     Find  DC  and  .:  BC.  Fig.  476 

6.  Find  one  side  of  the  regular  inscribed  duodecagon  if  the 
radius  of  the  circle  is  R. 

7.  Find  a  side  of  the  regular  circumscribed  hexagon  if  the  radius 
of  the  circle  is  1 ;  if  it  is  R. 

8.  Find  one  side  of  the  regular  inscribed  decagon  if  the  radius  of 
the  circle  is  1. 

Suggestion.     Let  s  represent  one  side  of  the  decagon.     Show  that  s 
may  be  obtained  from  the  equation  —  =  - — -  . 

9.  Find  one  side  of  the  regular  inscribed  decagon  if  the  radius 
of  the  circle  is  R. 

10.  Using  the  side  of  the  regular  inscribed  decagon 
found  in  Ex.  9  and  the  radius  of  the  circle  as  1,  find 
a  side  of  the  regular  inscribed  pentagon. 

Suggestion.     In  Fig.  477,  BC  is  a  mean  proportional 
between  DC  and  CE.     BE  can  be  found  from  BC  and  CE.       Fig.  477 


268  PLANE  GEOMETRY 

11.  Find  the  ratio  of  the  side  of  a  square  inscribed  in  a  circle 
to  the  side  of  an  equilateral  triangle  inscribed  in  the  same  circle. 

12.  ABCDEF  is  a  regular  hexagon.  X,  Y,  and  Z  are  the  mid- 
points of  AF,  BC,  and  DE  respectively.  Prove  that  XYZ  is  an 
equilateral  triangle  and  find  its  area  if  AB  =  20  in. 

STAR  POLYGONS 

293.  Fig.  474  shows  a  star  polygon  of  five  points  and 
Fig.  467  one  of  six  points.  There  are  one  or  more  regular 
star  polygons  related  to  each  of  the 
regular  convex  polygons.  They  can 
be  constructed  by  dividing  the  circle 
into  n  equal  parts  and  joining  each 
point  of  division  to  the  ^th  one  from 
it,  where  k  is   an  integer  greater  than 

one  and  less  than  -  •     Fig.  478  shows 

another  illustration.  (See  also  Fig. 
258.)  These  figures  have  been  studied 
wherever  geometry  has  been  studied.  They  abound  in  cut- 
glass  designs.  The  five-pointed  star  mentioned  in  the  note, 
§291,  Ex.  14,  and  the  six-pointed  star  mentioned  in  the 
aote,  §291,  Ex.  2,  are  of  special  importance. 

Exercise.  In  Fig.  478  the  circle  is  divided  into  16  equal  parts 
and  each  point  joined  to  the  7th  from  it.  The  following  questions 
apply  to  Fig.  478. 

1.  How  many  sets  of  equal  angles  are  there? 

2.  Find  the  number  of  degrees  in  one  angle  of  each  set. 

3.  Prove  that  AH  =  HB  =  BK  =  etc. 

4.  How  many  other  sets  of  equal  segments  are  there? 

5.  How  many  regular  polygons  of  16  sides  can  be  formed  by 
joining  corresponding  intersections?     Give  proof  for  each. 

6.  Can  regular  octagons  be  formed  by  joining  corresponding 
intersections?    How? 

7.  How  many  squares  are  in  the  figure? 


CHAPTER   XIII 

Measurement  of  the  Circle 

THE   CIRCUMFERENCE   OF  THE  CIRCLE 

DEFINITION 

294.  The  length  of  a  straight-line  segment  was  defined 
as  the  number  of  times  a  certain  straight-line  segment 
taken  as  a  unit  can  be  applied  to  the  segment  to  be  meas- 
ured. It  is  at  once  evident  that  we  cannot  measure  a  circle 
in  this  way.  We  shall,  however,  assume  that  the  circle  can 
be  measured  in  terms  of  a  straight-Hne  unit.  The  measure 
of  the  circle  is  called  its  length  or  its  circumference. 

GENERAL   METHOD 

295.  The  perimeter  of  an  inscribed  or  of  a  circumscribed 
polygon  of  many  sides  may  be  used  as  an  approximation  to 
the  length  of  the  circle. 

If  a  regular  hexagon  is  inscribed  in  a  circle  and  the  arcs 
between  its  vertices  are  bisected,  the  chords  joining  the 
points  of  division  form  a  regular  inscribed  polygon  of  12  sides. 
By  the  same  process  regular  polygons  of  24,  48,  96,  192 
sides  may  be  obtained,  and  so  on  indefinitely.  If  such  a 
succession  of  polygons  is  constructed,  one  is  soon  found 
which  can  with  difficulty  be  distinguished  from  the  circle. 
Instead  of  the  regular  inscribed  hexagon,  the  inscribed 
square  might  have  been  used  as  the  starting  point. 

Ex.  1.  The  perimeter  of  a  regular  inscribed  poly- 
gon is  less  than  the  perimeter  of  the  regular  polygon 
of  double  the  number  of  sides  inscribed  in  the  same 
circle  (Fig.  479). 

Fig.  479 

269 


270 


PLANE  GEOMETRY 


If  a  regular  hexagon  is  circumscribed  about  a  circle  and 
the  arcs  between  the  points  of  tangency  are  bisected,  the 
tangents  drawn  at  the  points  of  division  form  a  regular 
circumscribed  polygon  of  12  sides.  By  the  same  process, 
regular  circumscribed  polygons  of  24,  48,  96,  192  sides  may 
be  obtained,  and  so  on  indefinitely.  Just  as  in  the  series  of 
regular  inscribed  polygons,  so  in  this  series  of  regular  cir- 
cumscribed polygons,  a  polygon  is  soon  found  which  can 
with  difficulty  be  distinguished  from  the  circle.  The  start- 
ing point  of  this  series  might  have  been  the  circumscribed 
square  instead  of  the  regular  circumscribed  hexagon. 

The  perimeters  of  the  polygons  in  these  series  have  been 
computed.  The  computations  can  be  made  as  indicated  in 
the  next  two  problems. 

Ex.  2.  The  perimeter  of  a  regular  circumscribed 
polygon  is  greater  than  the  perimeter  of  the  regular 
polygon  of  double  the  number  of  sides  circum- 
scribed about  the  same  circle  (Fig.  480). 

Fig.  480 

296.  Problem  33.  To  find  a  side  of  a  regular  inscribed 
polygon  of  2n  sides,  given  a  side  of  a  regular  inscribed 
polygon  of  n  sides  and  the  radius  of  the  circle. 


Fig.  481 


Given  a  side  AC  (or  a)  of  a  regular  polygon  of  n  sides 
inscribed  in  a  circle  of  radius  r. 

Required  to  find  a  side  (x)  of  a  regular  inscribed  polygon 
of  2w  sides. 


MEASUREMENT  OF  THE  CIRCLE  271 

Analysis  (segments  are  lettered  as  shown  in  Fig.  481) : 
I.  To  find  X  in  terms  of  the  known  lengths  use 
x^  =  d^-\-{i  a)-,     a  is  known;  fi  must  be  found. 

TT    a>    ^   J  ^  J  I  ri^  known. 

II.  To  find  a,  use  a  =  r  — V    i  .  .     r        i 

[  y  must  be  found. 

III.  To  find  y,  use 

J"  =  r^  —  {\  ay.     a  and  r  are  known. 

Let  the  pupil  make  the  ntlmerical  computation  when  ^  C  is  a  side 
(1)  of  the  inscribed  sciuare,  (2)  of  the  regular  inscribed  hexagon.  Use 
for  the  radius  of  the  circle  (1)  \,  (2)  r.  The  following  formula  gives 
the  value  of  x  in  terms  of  r  and  a  :  jc=  V  2r'^  —  r  \'4r'-— a^. 

297.  Problkm  34.  To  find  a  side  of  a  regular  circum- 
scribed polygon  of  2n  sides,  given  a  side  of  a  regular  circum- 
scribed polygon  of  n  sides  and  the  radius  of  the  circle. 


Fig.  482 

Given  a  side  AB  (or  a)  of  a  regular  polygon  of  n  sides 
circiunscribed  about  a  circle  of  radius  r. 

Required  to  find  a  side  {%)  of  a  regular  circumscribed 
polygon  of  2w  sides. 
Analysis  (the  segments  are  lettered  as  shown  in  Fig.  482)  ; 
I.  To  find  X,  use  Th.  106. 

''  —        2  ^  a  and  r  are  known. 


y      2^'~2^  [>'tobe  found. 

II.  To  find  y,  use  y'^=-{\  ay-\-r-. 

Let  the  pupil  make  the  numerical  computation  when  AB  \s,  o.  side 
of  the  circumscribed  square.    Use  for  the  radius  of  the  circle  (I)  i ;  (2)  r. 

The  general  formula  is  x=  - — ; — ,-7  o  ■    ..  • 
2r+  V4r2-f  a-^ 


272 


PLANE   GEOMETRY 


PERIMETERS   OF  REGULAR   POLYGONS   INSCRIBED   IN  OR 
CIRCUMSCRIBED   ABOUT  A  CIRCLE  WITH  DIAMETER   ONE 

298.  The  methods  indicated  in  Probs.  33  and  34  have  been 
used  in   obtaining  the  results  given  in  the  tables  below. 

TABLE     I 


Number 

OF 

Sides 

Perimeter 

OF 

Inscribed  Polygon 

Perimeter 

OF 

Circumscribed  Polygon 

6 

3. 

3.464101 

12 

3.105828 

3.215390 

24 

3.132628 

3.159659 

48 

3.139350 

3.146086 

96 

3.141031 

3.142714 

192 

3.141452 

3.141873 

•  384 

3.141557 

3.141662 

768 

3.141583 

3.141610 

1536 

3.141590 

3.141597 

3072 

3.141592 

3.141594 

6144 

3.141592 

3.141593 

TABLE  II 


Number 

OF 

Sides 

Perimeter 

OF 

Inscribed    Polygon 

PiiRIMETER 
OF 

Circumscribed  Polygon 

4 

2.828427 

4.000000 

8 

3.061667 

3.313708 

16 

3.121445 

3.182597 

32 

3.136548 

3.151724 

64 

3.140331 

3.144118 

128 

3.141277 

3.142223 

256 

3.141513 

3.141750 

512 

3.141572 

3.141632 

1024 

3.141587 

3.141602 

2048 

3.141591 

3.141595 

4096 

3.141592 

3.141593 

8192 

3.141592 

3.141592 

MEASUREMENT  OF  THE  CIRCLE  273 

Table  I  shows  the  perimeters  of  a  series  of  inscribed  poly- 
gons and  of  a  series  of  circumscribed  polygons  that  start 
with  the  regular  inscribed  and  circumscribed  hexagon 
respectively.  Table  II  shows  the  perimeters  of  two  series 
that  start  with  the  inscribed  and  circumscribed  square 
respectively.  The  diameter  of  the  circle  is  1  in  each  case. 
Can  you  come  to  any  conclusion  by  comparing  results  ? 

In  higher  mathematics  it  is  proved  that  the  perimeters 
of  the  polygons  in  Tables  I  and  II  approach  a  definite  num- 
ber. This  number  cannot  be  found  exactly.  It  has  been 
named  w  (pi). 

COMPUTATION   OF  PERIMETERS   OF  REGULAR  POLYGONS 
INSCRIBED  IN  OR  CIRCUMSCRIBED  ABOUT  ANY  CIRCLE 

299.  Problem  35.  To  find  the  perimeters  of  regular 
polygons  inscribed  in  or  circumscribed  about  any  circle. 

Solution: 

I.  Let  d  and  d'  represent  the  diameters  of  two  circles. 
p  and  p'  represent  the  perimeters  of  two  regular 
polygons  of  the  same  number  of  sides  inscribed  in 
or  circumscribed  about  the  two  circles. 

II.  Then  ^  =  ^ (see  Th.  135  Cor.). 

p     a 

^,  =  d,whend'  =  l. 
P 
.'.  p  =  dp'  if  the  diameter  of  circle  about  p'  is  1. 
The  perimeter  of  a  regular  polygon  of  any  number  of  sides 
inscribed  in  or  circumscribed  about  a  circle  can  be  found 
from  the  equation  p  =  dp\   if  the  perimeter  of  a  similar 
polygon  inscribed  in  or  circumscribed  about  a  circle  of 
diameter  1  is  known. 

The  perimeters  of  polygons  inscribed  in  or  circumscribed 
about  any  circle  and  similar  to  those  in  Tables  I  and  II 
may  be  found  by  multiplying  the  diameter  of  the  given 
circle  by  the  proper  number  in  the  tables. 


274  PLANE   GEOMETRY 

LIMITING  VALUES  OF  PERIMETER  OF  INSCRIBED  AND 
CIRCUMSCRIBED   POLYGONS 

300.  As.  64.  The  limit  of  the  perimeters  of  a  series  of 
regular  polygons  inscribed  in  or  circumscribed  about  the 
same  circle  as  the  number  of  sides  is  increased  indefinitely 
is  the  same. 

This  limit  doe?  not  depend  upon  the  number  of  sides  of 
the  initial  polygon  nor  upon  the  method  of  increasing  the 
number  of  sides.     This  limit  is  ird. 

This  assumption  is  proved  in  higher  mathematics. 

LENGTH    OF   THE   CIRCLE 

301.  The  assumptions  in  §300  lead  us  to  the  following 
definition : 

The  length  or  the  circumference  of  a  circle  is  defined  as 
the  limit  of  the  perimeters  of  a  series  of  regular  polygons 
inscribed  in  or  circumscribed  about  a  circle  as  the  number 
of  sides  is  increased  indefinitely. 

The  perimeter  of  any  one  of  the  polygons  may  be  regarded 
as  an  approximation  to  the  length  of  the  circle. 

From  As.  64  and  the  definition  above,  we  have 
Theorem  137.    The  circumference  of  a  circle  of  diameter 
d  is  Tvd, 

If  c  is  the  circumference,  d  the  diameter,  r  the  radius, 
We  have  .*.  c=  ird  or  c  =  2Trr. 

Note.  The  number  tt  is  a  very  important  number  in  mathematics. 
It  is  an  irrational  number  (see  §198),  but  very  unHke  such  irrational 
numbers  as  V2  or  V3.  We  found  it  possible  by  means  of  a  straight- 
edge and  compasses  to  construct  a  straight-line  segment  to  represent 
V2  (Ex.  2,  §221).  The  ancient  Greeks  could  do  this.  They  failed, 
however,  in  the  attempt  to  construct  a  straight-line  segment  to  repre- 
sent TT.  In  modern  times  it  has  been  proved  that  it  is  impossible  to 
construct  such  a  segment  for  tt  by  the  use  of  straightedge  and  com- 
passes. It  can,  however,  be  done  by  means  of  an  instrument  called  an 
integraph,  invented  by  a  Pole  about  1878. 


MEASUREMENT  OF  THE  CIRCLE  275 

AREAS   OF   CIRCLES,   SECTORS,   AND   SEGMENTS 
THE  AREA   OF  THE   CIRCLE 

302.  As.  65.  The  limit  of  the  areas  of  a  series  of  regular 
polygons  inscribed  in  or  circumscribed  about  the  same  circle 
as  the  number  of  sides  is  increased  indefinitely  is  the  sam:'. 
This  limit  is  one-half  the  product  of  the  radius  of  the 
circle  and  its  circumference. 

This  assumption  is  proved  in  higher  mathematics. 

303.  We  shall  define  the  area  of  a  circle  as  the  limit  of 
the  areas  of  a  scries  of  inscribed  or  circumscribed  regular 
polygons  as  the  number  of  sides  is  increased  indefinitely. 

304.  From  §§302  and  303  we  have 

Theorem  138.  The  area  of  a  circle  is  one-half  the 
product  of  its  radius  and  circumference. 

Using  A  for  the  area  of  the  circle,  r  for  the  radius,  c  for 
the  circumference, 
A  =  y2cr 

=  H  r  '  2Trr=Trr^ since  c  =  27rr. 

Note.  We  know  that  tt  represents  a  definite  number,  although 
we  cannot  find  that  number  exactly.  We  have  learned  that  a  straight- 
line  segment  can  be  constructed  which  is  equal  to  the  circumference 
of  a  given  circle,  although  it  cannot  be  constructed  with  compasses  and 
straightedge.  We  will,  therefore,  assume  that  there  is  a  triangle, 
square,  or  rectangle  equivalent  to  any  given  circle,  although  these 
figures  cannot  be  constructed  with  straightedge  and  compasses. 

AREAS   OF   SECTORS 

305.  As.  66.  The  area  of  a  sector  has  the  same  ratio  to 
the  area  of  a  circle  of  which  it  is  a  part  as  the  angle  of  the 
sector  has  to  four  right  angles. 

Using  5  for  the  area  of  the  sector,  a  for  the  angle  of  the 
sector,  7rr-  for  the  area  of  the  circle, 

Solvingfor  5,  5  =  -^- 


270  PLANE   GEOMETRY 

AREAS   OF   SEGMENTS 

306.  A  circular  segment  is  a  figure  boimded  by  an  arc 
and  its  chord. 

The  relation  between  circular  segments  and 
sectors  may  be  shown  by  constructing  the  sector 
having  the  same  arc  as  a  given  circular  segment. 
It  is  at  once  evident  (Fig.  483)  that  the  area 
of  a  circular  segment  can  be  obtained  if  the  Fig.  483 
areas  of  the  associated  sector  and  triangle  can  be  obtained. 

RATIOS  AND   CIRCLES 

307.  The  following  theorems  follow  at  once  from  the 
formulae  :. 

Theorem  139.  The  ratio  of  the  circumference  to  the 
diameter  is  the  same  for  all  circles. 

If  c  =  wd,  then  -r  =  tt. 
a 

Theorem  140.    The  ratio  of  the  circumferences  of  two 

circles  equals  the  ratio  of  their  diameters  or  of  their  radii. 

Using  c  and  ci  for  the  circumferences  of  two  circles,  d  and 
di  for  their  diameters,  r  and  ri  for  their  radii,  we  have: 
c  _Ci  ^    c  _d 

d    di  "  Ci     di 

^.  d     2r       r  c       r 

Smce    -r  =  ?r~=~">  •'•~r  =  — 

di     2ri     n  c^     ri 

Theorem  141.  The  ratio  of  the  areas  of  two  circles 
equals  the  ratio  of  the  squares  of  their  radii  or  of  their 
diameters. 

Using  A  and  Ai  for  the  areas  of  two  circles,  r  and  n  for 
their  radii,  d  and  di  for  their  diameters,  we  have: 


A      7rr2 
Ai     Trn^ 

A  r' 
"  A^    n' 

r      d     r'~ 
Smce  -  =  j-»  —7, 
n    di    Ti 

dr 

.  A  d' 
"Ai    di" 

B. 

For  two  circles : 

c 

d 

; 

— 

C\ 

rfi 

c 

r 

: 

=  — 

Ci 

n 

A 

r2 

d" 

— - 

— 

Ai' 

u' 

d,' 

MEASUREMENT  OF  THE  CIRCLE  277 

SUMMARY  AND  SUPPLEMENTARY  EXERCISES 
308.     FORMULAE  OBTAINED  IN  CHAPTER  XIII 

A.  For  one  circle: 

c  =  2Trr  =  ird 
A  =  7rr2 
c 


C.  For  sectors: 
_  wr-a 

EXERCISES  INVOLVING  THE   MEASUREMENT   OF  THE 
CIRCLE 

309.  1.  Find  the  circumference  and  the  area  of  a  circle  if  the 
diameter  is  4  ft.;  6  ft.;  25  ft. 

2.  Find  the  radius  of  a  circle  if  the  circumference  is  10  ft. ; 
25  ft.;    63  ft. 

3.  Find  the  radius  of  a  circle  if  the  area  is  20  sq.  ft.;  56  sq.  ft.; 
96  sq.  ft. 

4.  If  a  wheel  is  3  ft.  in  diameter,  how  many  revolutions  will 
it  make  in  going  a  mile? 

5.  What  is  the  diameter  of  a  wheel  that  makes  250  revolutions 
in  traveling  one  mile? 

6.  A  locomotive  is  running  at  the  rate  of  40  mi.  per  hour.  If 
the  driving  wheels  are  80  in.  in  diameter  and  the  truck  wheels  are 
36  in.,  how  many  revolutions  per  minute  does  each  make? 

7.  The  72-in.  drive  wheel  of  a  locomotive  makes  250  turns  per 
minute.     What  is  the  rate  of  the  locomotive  in  miles  per  hour? 

8.  Two  pulleys,  one  36  in.  and  one  15  in.  in  diameter,  are 
connected  by  a  belt.  The  36-in .  pulley  makes  200  turns  per  minute. 
How  many  turns  does  the  15-in.  pulley  make  per  minute? 

9.  It  is  the  general  practice  not  to  permit  the  rim  of  a  fly 
wheel  to  have  a  greater  velocity  than  a  mile  a  minute.  How  many 
revolutions  per  minute  may  be  allowed  an  8-ft.  fly  wheel? 

19 


278 


PLANE   GEOMETRY 


10.  Fig.  48-4  shows  the  construction  of  certain  types  of  trefoils. 
Find  the  entire  area  if  one  side  of  the  equilateral 
triangle  is  5  ft. 

11.  Find  the  ratio  of  the  area  of  a  circle  to  the 
area  of  the  inscribed  square  and  to  the  area  of  the 


Fig.  484 


Fig.  485 


circumscribed  square. 

12.  One  side  of  the  equilateral  triangle  in  Fig. 
485  is  7  ft.  Show  how  the  shaded  portion  is  formed 
and  find  its  area. 

13.  Show  how  the  shaded  figure  in  the  square  in 
Fig.  486  is  formed  and  find  its  area  if  one  side  of  the 
square  is  6  in. 

14.  Show  how  the  four  leaf-shaped  figures  in 
Fig.  487  are  formed  and  find  the  sum  of  their  areas 
if  one  side  of  the  square  is  6  in. 

15.  If  the  equatorial  diameter  of  the  earth  is  7926 

mi.,  what  is  the  length  of  a  degree  of  longitude  at 

the  equator? 

Fig.  487 

16.  (a)  Construct  the  locus  of  the  center  of  a  circle,  radius  one- 
half  inch,  which  rolls  around  an  equilateral  triangle,  altitude  two 
inches,  (b)  Compute  to  two  decimals  the  area  inclosed  by  the 
locus  and  the  perimeter  of  the  locus. — College  Entrance  Examina- 
tion Board,  Plane  Geometry  Examination,  1914. 

17.  A  circle  of  radius  2  in.  rolls  around  a  square  whose  side  is 
4  in.  Construct  the  locus  of  the  center  of  the  circle,  and  find  to  two 
decimal  places  both  the  length  of  the  locus  and  the  area  inclosed 
by  it. — College  Entrance  Examination  Board,  Plane  Geometry 
Examination,  1915. 

18.  If  a  circle  is  constructed  on  each  of  the  sides  of  a  right 
triangle  as  diameter,  the  area  of  the  circle  on  the  hypotenuse 
is  equal  to  the  sum  of  the  areas  of  the  other  two  circles. 

19.  Find  by  geometrical  construction  the  diameter  of  a  pipe 
the  area  of  whose  cross-section  is  equal  to  the  sum  of  the  areas  of 
the  cross-sections  of  two  given  pipes. 


MEASUREMENT  OF  THE   CIRCLE 


279 


20.  A  12-in.  water  pipe  branches  into  three  equal  pipes  whose 
combined  capacity  is  the  same  as  that  of  the  12-in.  pipe.  If  the 
quantity  of  water  carried  depends  upon  the  area  of  the  cross-section 
of  the  pipes,  what  must  be  the  diameter  of  each  of  the  three  pipes? 

21.  Semicircles  are  constructed  on  the  sides 

of  a  right  triangle  as  shown  in  Fig.  488.     Show 

that  the  sum  of  the  areas  of  the  two  crescents  is 

equal  to  the  area  of  the  triangle. 
^  ^  Fig.  488 

Note.     This  problem  is  due  to  Hippocrates  (about  470  B.C.).     His 

solution  is  the  first  case  of  the  area  of  a  curvilinear  figure  proved  equal 

to  the  area  of  a  rectilinear  figure. 

22.  A  circular  grass  plot  12  ft.  in  diameter  is  cut  by  a  straight 
gravel  path  3  ft.  wide,  one  edge  of  which  passes  through  the  center 
of  the  plot.  What  is  the  area  of  the  remaining  grass  plot? — 
College  Entrance  Examination  Board,  Plane  Geometry  Examina- 
tion, 1911. 

23.  The  diameters  of  two  circular  pulleys  are  respectively  12  ft. 
and  2  ft.,  and  the  distance  between  their  centers  is  10  ft.  Find 
the  length  of  the  shortest  string  which  will  go  around  the  pulleys, 
correct  to  three  significant  figures. — College  Entrance  Examina- 
tion Board,  Plane  Geometry  Examination,  1910. 

24.  There  are  many  reasons  why  an  egg- 
shaped  sewer  is  more  satisfactory  than  a  circu- 
lar sewer.  The  cross-section  of  the  sewer  in 
Fig.  480  is  made  up  of  four  circular  arcs  A  By 
BC,  CD,  and  DA.  The  center  of  arc  AB  is 
F;  of  arc  yl  D  is  P;  of  arc  DC  is  X.  The  fig- 
ure i^  symmetric  with  respect  to  line  XY. 


Fig.  489 


If  ;?=  1 .2  ft.,  r=  .7  ft.,  ^  =  2.0  ft.,  Z  P  =  30°,  find  the  entire  circum- 
ference of  the  sewer.     AB  is  a,  semicircle. 

25.  Find  the  length  of  a  circular  railway  curve  of  radius  K 
mile  and  central  angle  30°. 

26,  The  rail  of  a  street-car  track  is  12  ft.  from 
the  curb.     In  rounding  a  square  corner  it  cannot 
be  closer  than  3  ft.  from  the  curb.    What  is  the 
radius  of  curvature  of  the  required  curve?     How        6 
far  back  must  the  curve  begin?     (Fig.  490.)  Fig.  490 


280 


PLANE  GEOMETRY 


27.  To  relieve  the  jolt   that  always  comes   in  passing   from 
a  tangent   to   a    sharp    curve,    many   railway 
curves  are  made  up  of  arcs  of  different  curva- 
ture.    Find  the  length  of  the  compound  curve 
in  Fig.  491. 

^O(±5^)=300  ft.     C  is  center  of  AM. 
CO  =  C'O  =  OD=150  ft.     C  is  center  of  NB. 
ZP  =  60°.     Z)  is  center  of  ikfiV.     The  figure 
is  symmetric  with  respect  to  OP. 

28.  Find  the  area  of  a  circular  segment  if  its  arc  is  60°;   45^ 

29.  In  Fig.  492,  the  arcs  AC  and  BC  are 
constructed  with  B  and  A  as  centers  respec- 
tively and  AB  as  radius.  Show  that  the  entire 
area  of  the  circular  figure  may  be  found  by 
subtracting  the  area  of  the  triangle  from  twice  ^ 
the  area  of  the  sector.     What  is  the  number  Fig.  492 

of  degrees  in  the  angle  of  the  sector?     Find  the  area  if  AB  =  8  ft. 

30.  The  figure  shown  in  Fig.  493  is  a  form  ^ 
frequently  used  in  church  window  designs.     Show 
how  it  is  constructed  and  find  its  area  if  ^J5  =  6  ft. 

31.  Find  the  area  of  the  ring  between  two  con- 
centric circles  if  the  radii  of  the  circles  are  5  in.  and  ^' 
3  in.  respectively.  Fig.  493 

32.  Find  a  formula   for   the  area  of  the   ring  between  con- 
centric circles  if   the  radii  are  R  and  r  respec- 
tively. • 

33.  Fig.  494  shows  two  concentric  circles. 
Show  how  to  find  the  segment  AB  so  that  the 
circle  constructed  on  AB  as  diameter  shall  have 
the  same  area  as  the  ring.  Fig.  494 

34.  The  circumferences  of  two  concentric  circles  differ  by  6  in. 
Compute  the  width  of  the  ring  between  the  two  circles  correct  to 
three  significant  figures. — College  Entrance  Examination  Board, 
Plane  Geometry  Examination,  1909. 

35.  Compare  the  area  of  a  circle  with  the  area  of  a  square  if 
the  perimeter  of  the  square  equals  the  circumference  of  the  circle. 


MEASUREMENT  OF   THE   CIRCLE 


281 


36.  Show  how  Fig.  405  is  formed  and  find  the 
area  of  HKLMFEH  and  of  HKLMFGH  if  one  side 
of  the  square  A  BCD  is  6  in.  ♦ 

37.  If  the  sum  of  the  radii  of  two  given  circles 

is  equal  to  the  radius  of  a  third  circle,  prove  that         Fig.  495 

the  circumference  of  the  third  circle  is  equal  to  the      From  an  old 

r    1         •  c  r     1  •  .     ,  Roman 

sum  of  the  circumferences  of  the  two  given  circles.        pavement 


38.  Construct  a  circle  whose  circumference  is 
equal  to  the  difTerence  between  the  circumferences 
of  two  given  circles. 

39.  Show  how  Fig.  490  is  formed  and  find  its 
area  if  a  side  of  the  hexagon  is  5  ft. 

40.  In  Fig.  497  the  diameter  of  the  circle  was 
divided  into  three  equal  parts.     Show  how  the  arcs  a 
shown  are  drawn  and  prove  that  they  divide  the 
circle  into  three  equal  parts. 


Fig.  496 


Fig.  497 


41.  Show  how  Fig.  498  is  formed, 
area  of  OABCD  if  05  =  5  ft. 


Find  the 


42.  Show  how  to  divide  a  circle  into  four  equal 
parts  by  the  arcs  of  circles  only.  Can  this  prob- 
lem be  solved  in  more  than  one  way? 


Note.  Ex.  42  is  said  to  be  a  problem  which  Napoleon  once  pro- 
posed to  his  staff.  The  figure  formed  by  dividing  the  circle  into  two 
equal  parts  (Fig.  498)  is  the  trademark  of  one  of  the  western  railroads. 


43.  In  Fig.  499,  C  is  the  mid-point  of  radius  ^0  of  OO.     CX 
is  perpendicular  to  AO  at  C.     Prove  that  the  circle 
with  O  as  center  and  OX  as  radius  has  half  the 
area  of  the  circle  with  radius  AO. 

44.  Show  how  to  divide  a  circle  into  three  equal 
parts  by  concentric  circles.  Fig.  439 


282  PLANE  GEOMETRY 

45.  It  is  desired  to  construct  a  half-mile  track.  The  start 
and  finish  are  to  be  straightways  intersecting  at  right  angles  at 
the  goal.  The  rest  of  the  track  is  to  be  an  arc  of  a  circle  tangent 
to  the  two  straightways.  Find  the  radius  of  the  arc  and  the  length 
of  the  arc  in  feet;  also  the  area  inclosed  by  the  track  in  acres. 
(Results  to  be  correct  to  two  decimals.) — College  Entrance  Exami- 
nation Board,  Plane  Geometry  Examination,  1914. 

APPROXIMATION  CONSTRUCTIONS 

310.  We  have  seen  that  numbers  can  be  found  which 
more  or  less  closely  approximate  the  value  of  tt,  although 
a  number  cannot  be  found  which  is  exactly  equal  to  ir. 
Similarly,  segments  can  be  constructed  whose  lengths  more 
or  less  closely  approximate  the  value  of  tf,  although  no 
segment  can  be  constructed  with  ruler  and  compasses  whose 
length  exactly  represents  the  value  of  tf.  The  following 
exercises  give  some  of  these  constructions.  They  are  used 
by  draftsmen  in  obtaining  the  development  of  cylinders, 
by  carpenters  in  obtaining  the  dimensions  of  veneers  for 
semicircular  heads  of  doors  and  windows,  and  by  mechani- 
cal tradesmen  generally.  The  computations  for  the  true 
lengths  of  the  segments  often  involve  considerable  geome- 
try, and  are  necessary  in  order  to  ascertain  the  degree  of 
approximation  obtained. 

Ex.  1.     In  Fig.  500,  ACB  is  a  semicircle  with  /.  e  £ 

AB  its  diameter.  AABD  is  equilateral,  with  AB  \ 
as  one  side.  EF  is  a  tangent  to  the  semicircle 
parallel  to  AB.  DA  and  DB  are  extended  to  meet 
the  tangent  EF  at  F  and  E  respectively.  Using 
AB  =  Q,  find  the  difference  between  the  length  of 
EF  and  of  the  semicircle  ACB.  Find  the  ratio  of 
this  difference  to  the  length  of  A  CB. 

Ex.  2.  In  Fig.  501,  ACB  is  a  semicircle  with  AB 
its  diameter.  CO ±  bisector  of  AB.  ZOCZ)  =  30°. 
Find  the  difference  between  the  length  oi  AD  and  of 


the  arc  AC,  using  AB  =  5.     Find  the  ratio  of  this  ^         o     n  b 
difference  to  the  length  of  the  arc  AC.  ^^^-  ^^^ 


MEASUREMENT  OF   THE   CIRCLE  283 

Ex.  3.     In  Fig.  502,  Z  ^0£  is  a  central  angle  of  45°  in  circle  O. 
OELAB.     Using  0£  =  8,  find  the  length  of  DE 
and    the    difference    between   WE -{-DE   and    the 
circumference.     Find   the   ratio  of  this  difference 
to  the  circumference. 

Ex.  4.     Inscribe    a   square    in    a   given   circle. 
Find  the  value  of  three  times  the  diameter  of  the  ' 

circle  plus  one-fifth  of  the  side  of  the  square,  using  the  diameter 
of  the  circle  as  d.  Find  the  difference  between  this  sum  and 
the  circumference  of  the  circle.  Find  the  ratio  of  this  difference 
to  the  circumference  of  the  circle.  (From  Ball's  Mathematical 
Recreations  and  Essays.) 

Ex.  5.  Determine  the  value  of  w  by  the  following  experiment: 
Wrap  a  paper  about  a  cylinder.  Stick  a  pin  through  the  over- 
lapped paper.  Measure  the  distance  between  the  two  pinholes 
and  the  diameter  of  the  cylinder  and  compute  w. 

311.  We  do  not  know  who  made  the  first  attempt  to  compare  the  area 

of  a  circle  with  the  area  of  a  square.     But  the  first  record  that  has  so 

far  been  found  is  on  an  Egyptian  papyrus  by  Ahmes  (about  1700  B.C.). 

He  says,   "Cut  off  }i  of  a  diameter  and  construct  a  square  on  the 

256 
remainder."    This  implies  that  the  area  of  the  circle  is  {%  d)^  or-^r'; 

81 

256 
that  is,  the  approximation  to  w  is  -^^j- =3. 16049 -|-.     This  is  a  much 

ox 

better  approximation  than  was  used  by  some  other  ancient  peoples. 
In  the  countries  of  Asia  3  was  commonly  used.  Both  3  and  33^^  are  found 
in  the  Bible  (I  Kings  7:  23;  Daniel  7:  25).  Archimedes  (287-212  B.C.) 
used  inscribed  and  circumscribed  polygons  of  96  sides  and  showed  that 
TT  was  less  than  Sl'j  and  greater  than  3  ^^f  i .  His  method  was  practically 
the  only  method  used  for  the  next  2,000  years  until  the  invention  of  the 
calculus  in  modern  times.  S}<i  is  an  approximation  close  enough  for 
most  practical  work.  In  India  a  value  appear^ before  the  Christian 
Era.  In  later  times  the  Hindus  used  3M,  Vl0  =  3.162,_and  3.1416. 
This  last  value  was  given  in  476  a.d.  The  Chinese  used  V 10  as  early  as 
the  second  century  a.d.  In  1610  a  European  published  a  result  correct 
to  35  decimal  places.  Since  the  invention  of  the  calculus  new  methods 
of  computation  have  been  devised.  In  1874  there  was  a  value  found 
showing  707  decimal  places.  The  first  97  digits  are  as  follows:  7r  = 
3.141.592653r)89793238462643383279502884197169399375105820974944 
5923078164062862089986280348253421 17  + . 


CHAPTER  XIV 

Maxima  and  Minima 

INTRODUCTORY 

312.  Of  all  geometrical  magnitudes  that  fulfill  a  given 
requirement,  that  which  is  the  greatest  is  called  the  maxi- 
mum; that  which  is  the  least  is  called  the  minimum. 

TRIANGLES 
AREAS  OF  TRIANGLES  WITH  TWO  SIDES  GIVEN 

313.  Theorem  142.  Of  all  triangles  having  two  given 
sides,  that  in  which  these  sides  are  perpendicular  to  each 
other  has  the  greatest  area. 


Hypothesis:     In  A  ABC  and  ABD  side  AB  is  common 
and  AC  =  AD.     ZCAB  =  1  rt.Z. 
Conclusion:    Area,  ABO  area,  ABD. 
Analysis  and  construction: 
I.  To  prove  area  ABO  area,  ABD,  prove  altitude  of 

AASO altitude  of  AABD. 
II.    .-.draw  DE  from  D±AB  and  prove  DE<AC. 
III.  To  prove  DE<AC,  prove  DE<AD. 
The  proof  is  left  to  the  pupil. 

284 


MAXIMA  AND  MINIMA  285 

Ex.  1.     Of  all  parallelograms  having  given  sides,  the  rectangle 
has  the  greatest  area. 

Ex.  2.     Of  all  parallelograms  having  given  diagonals,  which 
has  the  greatest  area? 

Ex.  3.     Of  all  triangles  having   a  given   base   and   a  given 
median  to  the  base,  which  has  the  greatest  area? 

MINIMUM   PERIMETERS 

314.  Theorem  143.  Of  all  triangles  having  the  same 
base  and  the  same  area,  the  isosceles  triangle  has  the  least 
perimeter. 


Hypothesis:     In  AABC  and  AABD  the  base  AB  is  com- 
mon and  area  ABC  =  a,resL  ABD.     AABC  is  isosceles. 
Conclusion:     Perimeter  ABC  < perimeter  ABD. 
Analysis  and  construction: 

I.  To  prove  per.  ABC  <^er.  ABD,  prove 

AC+CB<AD^-DB. 
II.    .'.  extend  AC,  making  CE  —  CB.     Join  DE  and  DC 
AE<AD-^DE. 


and  prove    ,  ^^^^^ 

III.  To  prove  DE  =  DB,  prove  ACDE  ^ACDB. 

IV.  To  prove  ACDE  ^ACDB,  prove  Z 1  =  Z  2. 
V.  To  prove  Z 1  =  Z  2,  prove  CD  \\  AB. 

VI.  To  prove  CD\\AB,  draw  a  perpendicular  from  D  to 
AB  and  prove  the  figure  formed  a  parallelogram. 

The  proof  is  left  to  the  pupil. 


286 


PLANE  GEOMETRY 
MAXIMUM   AREAS 


315.  Theorem  144.  Of  all  triangles  having  the  same 
base  and  equal  perimeters,  the  isosceles  triangle  has  the 
greatest  area. 


\ 

0^ 

^ 

X 

B 

Fig. 

£05 

Hypothesis:  In  AABC  and  AABD  the  base  AB  is 
common,  and  perimeter  A5C  =  perimeter  ADB.  AABC 
is  isosceles. 

Conclusion:    Area  yl^C^area  ABD. 

Analysis  and  construction: 

I.  To  prove  area  ABC > area  ABD,  prove  the  altitude 

of  AABOthe  altitude  of  AABD. 
II.    .*.  draw  CX  and  DY±AB  from  C  and  D  respectively; 
draw  DE\\AB  from  D,  intersecting  CX  at  E,  and 
prove  that  E  lies  between  C  and  X\  that  is,  prove 
CX>EX. 

III.  To  prove  CX> EX,    join    EA    and    EB    and   prove 

AOAE. 

IV.  To  prove  AOAE,  prove  AC +CB> A E+EB. 
V.  To  prove  AC+CB>AE-\-EB,  prove 

AD+DB>AE-\-EB. 
VI.  To  prove  AD-{-DB>AE+EB,  prove  area  AEB  = 

area  ylD5. 
The  proof  is  left  to  the  pupil. 

Suggestion.  Show  that  AEB  is  an  isosceles  triangle.  Since  area 
AEB=avea,ADB,  AD+DB>AE+EB. 


MAXIMA  AND  MINIMA  287 

POLYGONS  IN   GENERAL 

PRELIMINARY  THEOREM 
316.  Theorem  145.     Of  all  polygons  having  all  sides  but 
one  equal  to  given  segments  taken  in  order,  that  with  the 
greatest  area  can  be  inscribed  in  a  semicircle  with  the 
undetermined  side  as  diameter. 


Fig.  508 

Hypothesis:  The  polygon  ABCDEF  is  the  maximum 
polygon  that  can  be  formed  with  all  sides  but  one  equal  to 
the  segments  AB,  EC,  CD,  DE,  and  EF  taken  in  the  order 
given. 

Conclusion:  ABCDEF  can  be  inscribed  in  a  semicircle 
with  ^  F  as  diameter. 

Analysis  and  construction: 

I.  To  prove  that  ABCDEF  can  be  inscribed  in  a  semi- 
circle with  A  Fas  diameter,  prove  that  any  vertex, 
as  D,  lies  on  the  semicircle. 
11.  To  prove  that  D  Hcs  on  the  semicircle,  join  DA  and 

DF  and  prove  /LADF=^l  rt.*Z. 
III.  To  prove  AADF^l  rt.  Z,  prove  area  ADF  a  maxi- 
mum. 

The  proof  is  left  to  the  pupil. 

Suggestion  for  step  III.  Use  indirect  proof.  If  area  ADF  is  not 
a  maximum,  we  may  increase  or  decrease  A  ADF  by  sliding  points 
A  and  /^  along  the  line  XFuntilarca/lD/''is  a  maximum.  If,  as  points 
A  and  Fmove  along  line  .Y  F,  figures  A  BCD  and  DEFrviwixxn  unchanged, 
the  area  of  ABCDE  would  be  increased,  which  is  contrary  to  tlie 
hypothesis. 


288 


PLANE   GEOMETRY 
TEST  FOR   MAXIMUM   AREAS 


317.  Theorem  146.  Of  all  polygons  that  have  their 
sides  equal  respectively  to  given  segments  taken  in  order, 
that  which  can  be  inscribed  in  a  circle  has  the  greatest  area. 


Hypothesis:  The  polygon  ABODE  is  inscribed  in  a  circle 
and  polygon  A'B'C'D'E'  cannot  be  inscribed  in  a  circle. 
AB  =  A'B\  BC  =  B'C\  CD  =  CD\  etc. 

Conclusion:    Area  ABODE  >  area  A'B'C'D'E'. 

Analysis  and  construction: 

To   prove   area   ABODE  >  area  A'B'O'D'E', 
Draw  diameter  AX. 
Join  OX  and  DX. 
Construct    AD'X'O' ^ADXO. 
Compare  area  ABOX  with  area  A'B'O'X'  and 

area  AEDX  with  area  A'E'D'X'. 
Subtract  area  DXO  from  area  ABOXDE  and 
area  D'X'C  from  area  A'B'0'X'D'E\ 

The  proof  is  left  to  the  pupil. 

Discussion.  Might  one  part  of  ABODE  have  the  same 
area  as  the  corresponding  part  of  A'B'O'D'E'l  Could  both 
parts  of  ABODE  have  the  same  area  as  the  corresponding 
parts  of  A'B'O'D'E'l    Why? 


MAXIMA  AND  MINIMA  289 

THE   MAXIMUM   AREA   WITH   GIVEN   PERIMETER 

318.  Theorem  147.  Of  all  polygons  with  a  given  perime- 
ter and  a  given  number  of  sides,  that  with  the  maximum 
area  is  regular. 


Fig.  508 

Suggestion.  To  prove  P  regular,  prove  that  it  can  be  inscribed  in  a 
circle  and  is  equilateral.  To  prove  AC=CB,  prove  AABC  a,  maxi- 
mum by  indirect  proof. 

THE   MINIMUM   PERIMETER   WITH   GIVEN   AREA 
319.  Theorem  148.     Of  all  polygons  with  the  same  area 
and  the  same  number  of  sides,  the  regular  polygon  has  the 
least  perimeter. 


Fig.  509 

Analysis  and  construction: 
I.  To  prove  per.  P<per.  P\  compare  P  and  P'  with  a 

third  polygon. 
II.    .*.  construct  the  regular  polygon  Q  with  the  same 
number  of  sides  and  same  perimeter  as  P'  and 
prove  per.  P<per.  Q. 

III.  To  prove  per.  P<per.  Q,  prove  area  P<area  Q. 

IV.  To  prove  area  P<area  Q,  prove  area  P'<area  Q. 
The  proof  is  left  to  the  pupil. 


290  PLANE   GEOMETRY 

REGULAR  POLYGONS 

THE  MAXIMUM   AREA  WITH   GIVEN   PERIMETER 

320.  Theorem  149.  Of  all  regular  polygons  with  a  given 
perimeter,  the  one  with  the  maximum  area  has  the  greatest 
number  of  sides. 


Analysis  and  construction: 

I.  To  prove  area  P>area  P\  compare  P  and  P'  with  a 

third  polygon. 
II.    .*.  join  E,  any  point  in  AB,  with  C  and  construct 
ACDE^AACE   and    prove    (1)    area   P'=area 
BCDE;    (2)  area  P>area  BCDE. 
III.  To  prove  area  P>area  BCDE,  prove  per.  P  =  per. 
BCDE. 

Discussion  and  conclusion:  In  the  same  manner  it  can 
be  proved  that  a  regular  polygon  of  five  sides  has  a  greater 
area  than  a  square  of  same  perimeter  and  that  a  regular 
hexagon  has  a  greater  area  than  a  regular  pentagon  of  same 
perimeter,  and  so  on. 

321.  Since  the  area  of  a  circle  has  been  defined  as  the 
limit  of  the  areas  of  a  series  of  regular  inscribed  polygons 
as  the  number  of  sides  is  increased  indefinitely,  we  have: 

As.  67.  The  area  of  a  circle  is  greater  than  the  area  of 
any  polygon  of  equal  perimeter. 

In  higher  mathematics  we  prove : 

As.  68.  Of  all  figures  having  the  same  perimeter,  the 
circle  has  the  maximum  area. 


MAXIMA  AND   MINIMA 


291 


THE   MINIMUM   PERIMETER   WITH   GIVEN   AREA 

322.  Theorem  150.  Of  all  regular  polygons  with  the 
same  area,  that  having  the  greatest  number  of  sides  has  the 
least  perimeter. 


Fig.  511 

Analysis  and  construction: 
I.  To  prove  per.  P<per.  P',  compare  P  and  P'  with  a 
third  polygon. 
II.   .*.  construct  the  regular  polygon  Q  with  the  same 
number  of  sides  as  P'  and  same  perimeter  as  P  and 
prove  per.  P'  >  per.  Q. 
III.  To  prove  per.  P'>per.  Q,  prove  area  P'>area  Q. 
The  proof  is  left  to  the  pupil. 

323.  Since  the  circumference  of  a  circle  has  been  defined 
as  the  limit  of  the  perimeters  of  a  series  of  regular  inscribed 
polygons  as  the  number  of  sides  is  increased  indefinitely, 
we  have: 

As.  69.  The  perimeter  of  a  circle  is  less  than  that  of  any 
polygon  of  the  same  area. 

In  higher  mathematics  we  prove : 

As.  70.  Of  all  figures  having  the  same  area,  the  circle 
has  the  minimum  perimeter. 

Note.  As.  70  has  an  important  application  in  engineering.  The 
flow  of  water  in  an  aqueduct  or  sewer  is  checked  by  the  friction  of  the 
water  on  the  walls.  The  friction  is  proportional  to  the  perimeter  of 
the  cross-section.  It  is  desirable  to  keep  this  perimeter  as  small  as 
possible.  The  circle,  then,  is  the  best  form  to  meet  this  condition. 
In  case  it  is  not  desirable  because  of  expense  and  other  considerations 
to  use  a  circle,  regular  polygons  are  often  used. 


NOTES  ON  ARITHMETIC  AND  ALGEBRA 
FRACTIONS 

324.  The  following  fundamental  law  of  fractions  under- 
lies all  operations  that  involve  fractions. 

Multiplying  or  dividing  numerator  and  denominator  of 
a  fraction  by  the  same  number  does  not  alter  the  value  of 
the  fraction. 

A.  The  sum  or  difference  of  two  or  more  fractions  that 
have  a  common  denominator  is  the  sum  or  difference  of  the 
numerators  divided  by  the  common  denominator. 

Two  or  more  fractions  that  have  not  a  common  denomi- 
nator must  be  reduced  to  a  common  denominator  before 
adding  or  subtracting.  To  reduce  fractions  to  a  common 
denominator,  apply  the  fundamental  law  given  above. 

Add  and  subtract  the  following: 

1    ^  +  ^ 
^'  12^18 

^'  24^36     6 

B.  The  product  of  two  fractions  is  the  product  of  the 
numerators  divided  by  the  product  of  the  denominators. 
Where  possible,  divide  numerator  and  denominator  by  com- 
mon factors. 

Multiply  the  following: 

16     15  be  a  —  b       x — 1 

292 


b      c 

5.4  ^+f: 

y     4v- 

4.  ±+l_± 
be     ac    ab 

^-  :.-hl     X-: 

NOTES  ON  ARITHMETIC  AND  ALGEBRA        293 

C.  The  quotient  of  one  fraction  divided  by  a  second  is 

the  product  of  the  first  multiplied  by  the  reciprocal  of  the 
second. 

Divide  the  following: 
,.  14,33  2.  ?^^6a6  3.  "'-"*  •  "-* 


20    25  8*3     —  "•  Uab^         2a 

ROOTS 
325.  The  rule  for  square  root  is  based  on  the  algebraic 
formula  {a-^by  =  a^-\-2ab-}-b\    Notice  that  a^-{-2ab-\-b^  may 
be  written  a^-\-b{2a-{-b).     The  method  is  illustrated  below: 


Illustration  1.     Find    V 694. 563 
694.56,3     |26.3 
4_ 
2(20)  =  40    294 
40+6  =  46    276 


2(260)  =  520       1856 
520+3  =  523       1569 


287 

In  the  illustration  above,  notice: 

(1)  The  number  was  divided  into  periods  of  two  figures 
each,  counting  to  the  left  and  to  the  right  from  the  decimal 
point. 

(2)  The  largest  square  under  6  is  4.  The  4  was  sub- 
tracted from  6  and  the  next  period  annexed.  This  gave  a 
remainder  of  294.  The  square  root  of  4,  or  2,  was  written 
as  the  first  figure  in  the  root. 

(3)  A  zero  was  placed  after  the  2,  making  20.  The  20 
was  doubled,  making  40.  The  40  is  used  as  a  trial  divisor 
for  the  remainder  294.  The  next  figure  of  the  root  is  either 
6  or  7.  The  6  is  added  to  the  40,  making  46.  The  46  is 
multiplied  by  6,  giving  276.  The  276  is  subtracted  from 
294,  leaving  18.     The  next  period  is  annexed,  giving  1856. 

(4)  The  process  above  is  repeated  at  each  step  of  the 
work;  thus,  a  zero  is  placed  after  26  and  the  result  doubled, 
giving  the  520.     The  work  is  then  continued  as  above. 

20 


294  PLANE  GEOMETRY 

In  general  we  may  say:  Annex  a  zero  to  the  part  of  the 
root  already  found  and  double  the  result.  To  this  result 
add  the  next  figure  of  the  root.  Multiply  the  result  by  the 
last  figure  of  the  root  found. 

Show  that  this  statement  may  be  regarded  as  a  trans- 
lation of  b{2a+b)  in  the  formula  {a-\'b)^  =  a^+b{2a-hh). 

Find  the  square  root  of  the  following: 

1.  1369  4.  106276  7.  6 

2.  3744  5.  3  8.  15 

3.  2304  6.  5  9.  7 

Because  of  the  frequent  occurrence  of  the  square  roots 
of  2  and  3  in  geometry  work,  the  application  of  the  following 
law  should  be  noted : 

326.  The  square  root  of  a  product  is  the  product  of  the 
square  roots  of  the  factors. 

Illustration  2.     36  =  4  X  9     /.  V36  =  Vi  X  V9 

This  law  is  used  most  conveniently  for  inexact  -square 
roots  when  one  factor  is  a  perfect  square. 

Illustration  3^    18  =  9X2     .*.  V 18  =  V9  X  V2  =  3  V2 
Notice  that    V2  occurs  when  the  side  of  a  square  and  n 
diagonal  of  the  square  are  used  in  the  same  exercise. 

Illustration  4.     12  =  4X3      .-.  Vr2  =  Vi  X  V3  =  2  V3 

Notice  that   V3  occurs  when  the  side  of  an  equilateral  tri- 
angle and  its  altitude  occur  in  the  same  exercise. 

Illustration  5.     20  =  4  X  5     /.  V20  =  Vi  X  V5  =  2  VS 
The  V5  occurs  in  connection  with  the  regular  decagon  and 
pentagon. 

Find  the   value   of   the   following    correct   to   three   decimal 
places.     Apply  the   law   given   above. 

1.  V8  4.    Vl08      7.    Vl28       10.    Vl50       13.    V54 

2.  Vl8        5.    V32         8.    V75         11.    Vl25       14.    V45 

3.  V27        6.    V80        9.    V320       12.    V98         15.    Vl80 


NOTES  ON  ARITHMETIC  AND  ALGEBRA  295 


16.    V20 

18.    V48 

20.    V50      22.    V288 

24.    V243 

17.    V96 

19.    V72 

21.    V300    23.    Vl62 

25.    V242 

327.    The  square  root  of  a  fraction. 

A.  If  the  denominator  is  a  perfect  square:  Find  the 
square  root  of  the  numerator  and  of  the  denominator  sepa- 
rately and  divide  the  first  result  by  the  second. 

144     12 


Illustration  6.     _ 

625    25 


Illustration  7.    -4/-  =  3/2  V3 


B.  If  the  denominator  is  not  a  perfect  square,  two  methods 
are  suggested: 

(1)  The  fraction  may  be  reduced  to  a  decimal  and  the 
square  root  of  the  result  found. 

(2)  Numerator  and  denominator  may  be  multiplied  by 
some  number  that  will  make  the  denominator  a  perfect 
square'  and  method  A  above  used. 

Illustration  8.     To  find  \'^  either 

1.  Find  square  root  of  .333333-}-;  or 

2.  Write  H  =  %  and  use  Vz  Vs. 

Find  the  value  of  the  following  correct  to  three  decimal  places: 


■•n's  "-vt   Wf  -Vi- 

WE    '■<}    «<l  "-Vl 

'■4    'Vl    'Wfs  'Wl 

Wl    Ul    "V!  "-Vl 

17-  -v/"  to  four  decimal  places. 


27 


29G  PLANE'  GEOMETRY 

EQUATIONS 

328.  The  method  of  solving  linear  equations  is  illustrated 
below : 

Illustration  I.     Solve  for  :»;:  — -. 2  =  4 ^ 

.    Multiply  both  sides  by  the  L.  C.  M.  of  the 

denominators 3(3a:-fl) -24  =  48 -4(:»:-hl) 

Performing  multiplications Ox+S— 24  =  48— 4x— 4 

Combining  terms 9jc-21=44— 4;c 

Add +21  and +4^  to  each  side 13a;  =  65 

Divide  both  sides  by  13 x  =  b 

Solve  the  following  equations: 

1    ^  =  ^-  3.      "^  3 


3  5  x-b     2.r+7 

c-7_3+2a;_^  ^    3-4:y    5-2x^        x 

7  4  '83  4 


329.  If  an  equation  contains  both  the  first  and  the  second 
powers  of  the  unknown,  two  methods  of  solution  are 
suggested. 

A.  The  equation  may  be  solved  by  factoring. 
Illustration  2.     Solve  for  a;:  x'^  —  x  =  20 

a;2-x-20  =  0 
(:r-5)  (jc+4)=0 
x-b  =  0  :r+4  =  0 

a;  =  5  x  =  ^ 

Notice  that  to  solve  an  equation  by  factoring,  one  member 
of  the  equation  must  be  zero. 

B.  The  equation  may  be  solved  by  completing  the  square. 

Illustration  Z.    Solve  for  ac:  Zx^-bx  =  '7     .     .     .     (1) 

bx    7  I 

Divide  both  sides  by  3    .      .      .      .      .     ^''--3=3"     •      •      •     (2) 

Add  the  square  of  {}i  •  %)  to  each 
side x^-^^-Oi  '  %)^  =  |+|      .     .     (3) 


NOTES  ON  ARITHMETIC  AND  ALGEBRA        297 

Take  the  square  root  of  each  side 

5  Vl09  

of  the  equation    ....     .t-^  =  ±  — ^=  ±3-^  Vl09  .     (4) 

5     ^10.44 
x-^=±-j-      ....     (5) 

5     10.44  5     10.44 

*"6"^     6  *~6         6 

^15.44  5.44 

6  6 

=  2.57+  =-.90+ 

Notice  in  (3),  (>i  •  H)^  is  added  to  the  left  side  to  make  the 
left  side  a  perfect  square.  It  is  added  to  the  right  side  to 
preserve  the  balance  of  the  equation.  (H  •  H^y  or  ^He,  is 
obtained  by  squaring  half  the  coefficient  of  x.  Notice  that 
in  step  (2)  the  equation  is  divided  by  3  to  make  the  first 
term  ic^,  which  is  a  perfect  square. 
Solve  the  following  equations: 

1.  a;2+3:c=18  3.  3:^2-ll:x:  =  2 

2.  2x''-x  =  \b  4.  2x''-\-bx  =17 

330.  To  solve  a  system  of  equations  consisting  of  two 
equations  containing  two  unknowns,  eliminate  one  of  the 
unknowns  and  solve  the  resulting  equation  for  the  other. 

A.  When  both  equations  are  of  the  first  degree,  eliminate 
by  addition  or  subtraction. 

5:r-4y  =  6.5 


Illustration  Ai.     Solve  for  aj  and  y:   ,-    ,  -       oo  or 

5:c-4y  =  6.5  (1) 

7jc+53;  =  38.25 (2) 

35;c-28y  =  45.5 (1)  X  7 

35:c+25>'=  191.25 (2)  X  5 

-53y= -145.75 Subtract  the  third 

^  =  2.75  equation  from  the  second 

Notice  that  x  may  be  found  by  multiplying  equation  (1) 

by  5  and  equation  (2)  by  4  and  adding  the  results  or  by 

substituting  2.75  for  y  in  either  equation  (1)  or  (2)  and 

solving  the  result  for  x. 


298 


PLANE   GEOMETRY 


B.  When  one  of  the  equations  is  of  the  first  degree  and 
one  of  the  second,  solve  the  first-degree  equation  for  one  of 
the  unknowns  in  terms  of  the  other  unknown  and  substitute 
in  the  other  equation. 

\x  -hy  =8    .      .      . 
|x2+3;2  =  34  .      .      . 

x=8—y    . 

64-163;+2/  =  34 
2^2-16^+30  =  0 


Illustration  5.     Solve  for  x  and  y: 

Solve  (1)  for  x, 
Substitute  8  —  3;  for  :v  in  (2) 


(1) 
(2) 
(3) 


y^-8y+15  =  0 
(y-5)  (3'-3)=0 
y  =  D  and  y  =  3 
To  find  X,  substitute  the  values  of  y  in  (3) : 


y= 

5 

y- 

=  3 

x=8-y                           x=8-5 

=8-5                              =8-3 

=  3                                     =5 

\ x=3             ^ x=5 

The  solutions  are  i         ^             in 

[y==5          [y=^ 

Solve  the  following  systems  for  x  and  y : 

1.  2x+33;=16                2.  bx-2y  =  U               3.  2x^-j-y^  =  57 

5.r-2>;  =  21                               xy  =  20                         x-y=l 

TABLES 

331.     TABLE    OF   SQUARE   ROOTS 

0 

1 

2 

3 

4 

5          6 

7 

3 

9 

0 

0 

1.000 

1.414 

1.732 

2.00c 

2.236  2.449 

2.645 

2.828 

3.000 

1 

3.162 

3.316 

3.464 

3.605 

3.741 

3.872 

4.000 

4.123 

4.242 

4.358 

2 

4.472 

4 .  582 

4.690 

4.795 

4.898 

5.000 

5.099 

5.196 

5.291 

5.385 

3 

0.477 

5.567 

5.656 

5.744 

5.830 

5.916 

6.000 

6.082 

6.164 

6.244 

4 

6.324 

6.403 

6.480 

6.557 

6.633 

6.708 

6.782 

6.855 

6.928 

7.000 

5 

7.071 

7.141 

7.211 

7.280 

7.34S 

7.416 

7.483 

7 .  549 

7.615 

7.681 

G 

7 .  745 

7.810 

7.874 

7.937 

S.OOO 

8.062 

8.124 

8.185 

8.246 

8.306 

7 

8.366 

8.426 

8.485 

8.544 

8.602 

8.660 

8.717 

8.774 

8.831 

8.888 

8 

8.944 

9.000 

9.055 

9.110 

9.165 

9.219 

9.273 

9.327 

9.380 

9.433 

9 

9.486 

9.539 

9.591 

9.643 

9.695 

9.746 

9.797 

9.848 

9.899 

9.949 

NOTES  ON   ARITHMETIC  AND  ALGEBRA        299 
332.     TABLE  OF  SINES,  COSINES,  AND  TANGENTS 


Deg. 

Sine 

Cosine 

Tangent 

Deg. 

Sine 

Cosine 

Tangent 

1 

.017 

.999 

.017 

46 

.719 

.695 

1.036 

2 

.035 

.999 

.035 

47 

.731 

.682 

1.072 

3 

.052 

.999 

.052 

48 

.743 

.669 

1.111 

4 

.070 

.998 

.070 

49 

.755 

.656 

1.150 

5 

.087 

.996 

.087 

50 

.766 

.643 

1.192 

6 

.105 

.995 

.105 

51 

.777 

.629 

1.235 

7 

.122 

.993 

.123 

52 

.788 

.616 

1.280 

8 

.139 

.990 

.141 

53 

.799 

.602 

1.327 

9 

.156 

.988 

.158 

54 

.809 

.588 

1.376 

10 

.174 

.985 

.176 

55 

.819 

.574 

1.428 

11 

.191 

.982 

.  194 

56 

.829 

.559 

1.483 

12 

.208 

.978 

.213 

57 

.839 

.545 

1  540 

13 

.225 

.974 

.231 

58 

.848 

.530 

1.600 

14 

.242 

.970 

.249 

59 

.857 

.515 

1.664 

15 

.259 

.966 

.268 

60 

.866 

.500 

1.732 

16 

.276 

.961 

.287 

61 

.875 

.485 

1.804 

17 

.292 

.956 

.306 

62 

.883 

.469 

1  881 

18 

.309 

.951 

.325 

63 

.891 

.454 

1  963 

19 

.326 

.946 

.344 

64 

.899 

.438 

2.050 

20 

.342 

.940 

.364 

65 

.906 

.423 

2.144 

21 

.358 

.934 

.384 

66 

.914 

.407 

2.246 

22 

.375 

.927 

.404 

67 

.921 

.391 

2.356 

23 

.391 

.921 

.424 

68 

.927 

.375 

2.475 

24 

.407 

.914 

.445 

69 

.934 

.358 

2  605 

25 

.423 

.906 

.466 

70 

.940 

.342 

2.747 

2d 

.438 

.899 

.488 

71 

.946 

.326 

2.904 

27 

.454 

.891 

.510 

72 

.951 

.309 

3.078 

28 

.469 

.883 

.532 

73 

.956 

.292 

3.271 

29 

.485 

.875 

.554 

74 

.961 

.276 

3.487 

30 

.500 

.866 

.577 

75 

.966 

.259 

3.732 

31 

.515 

.857 

.601 

76 

.970 

.242 

4.011 

32 

.530 

.848 

.625 

77 

.974 

.225 

4.331 

33 

.545 

.839 

.649 

78 

.978 

.208 

4.705 

34 

.559 

.829 

.675 

79 

.982 

.191 

5.145 

35 

.574 

.819 

.700 

80 

.985 

.174 

5.671 

36 

.588 

.809 

.727 

81 

.988 

.156 

6.314 

37 

.602 

.799 

.754 

82 

.990 

.139 

7.115 

38 

.616 

.788 

.781 

83 

.993 

.122 

8.144 

39 

.629 

.777 

.810 

84 

.995 

.105 

9.514 

40 

.643 

.766 

.839 

85 

.996 

.087 

11.430 

41 

.656 

.755 

.869 

86 

.998 

.070 

14.301 

42 

.669 

.743 

.900 

87 

.999 

.052 

19.081 

43 

.682 

.731 

.933 

88 

.999 

.035 

28.636 

44 

.695 

.719 

.966 

89 

.999 

.017 

57.290 

45 

.707 

.707 

1.000 

- 

300  PLANE   GEOMETRY 

333.    Units  of  Length 

English 
12  inches  (in.)  =  1  foot  (ft.) 
3  feet  =  1  yard  (yd.) 
5J  yards  =  1  rod  (rd.) 
320  rods  or  5280  ft.  =  1  mile  (mi.) 

Metric 
10  centimeters  (cm.)  =  1  decimeter  (dm.) 
10  decimeters  =  1  meter  (m.) 
1000  meters  =  1  kilometer  (km.) 

1  meter  =  39. 37  in. 
1  kilometer  =  .  62  of  a  mile 

1  foot  =  30.48  centimeters 
1  mile  =1.6093  kilometers 

334.    Units  of  Surface 

English 
144  square  inches  (sq.  in.)  =  1  square  foot  (sq.  ft.) 
9  square  feet  =  1  square  yard  (sq.  yd.) 
30M  square  yards  =  1  square  rod  (sq.  rd.) 

160  square  rods  =  l  acre  (A.) 
4840  square  yards  =  1  acre  (A.) 

640  acres  =  1  square  mile  (sq.  mi.) 

Metric 
100  square  centimeters  =  1  square  decimeter 
100  square  decimeters  =  1  square  meter 


OUTLINE  SUMMARY 

PARALLELS   AND   PERPENDICULARS 
Tests  for  parallels : 

Two  straight  lines  in  the  same  plane  are  parallel  if; 

1.  The  alt.  int.  angles  are  equal §61,  Th.  0 

2.  The  corresponding  angles  are  equal §63,  Th.  10 

3.  The  int.  angles  on  the  same  side  of  the  transversal 

are  sup §63,  Th.  11 

4.  They  are  perpendicular  to  the  same  line §63,  Th.  12 

5.  They  are  parallel  to  the  same  line §66,  Th.  13 

A  line  is  parallel  to  one  side  of  a  triangle  if: 

6.  It  passes  through  the  mid-points  of  the  other  two 

sides : §114,  Th.  48 

7.  It  divides  the  other  two  sides  proportionally 

§20o,  Th.  100  and  Cor. 
A  line  is  parallel  to  the  bases  of  a  trapezoid  if: 

8.  It  passes  through  the  mid-points  of  the  legs §118,  Th.  51 

Construction  of  parallels §64,  Prob.  6 

Tests  for  perpendiculars: 

1.  If  a  ray  starts  from  a  point  in  a  straight  hne §18 

2.  A  line  perpendicular  to  one  of  two  parallels §70,  Th.  17 

3.  A  line  tangent  to  a  circle  is  perpendicular  to .  §145,  Th.  69 

4.  If  any  two  circles  intersect  the  line  of  centers. . ,  .  §151,  Th.  73 

5.  If  two  equal  circles  intersect  the  line  of  centers  and 

the  common  chord §152,  Th.  74  Cor. 

6.  An  angle  inscribed  in  a  semicircle §162,  Cor.  II 

Construction  of  a  perpendicular: 

1.  To  a  line  from  a  point  in  the  line.  §43,  Prob.  3;  §167,  Prob.  9 

2.  To  a  line  from  a  point  not  in  the  line 

§44,  Prob.  4;  §167.  Prob.  10 

3.  To  a  segment  bisecting  the  segment §45,  Prob.  5 

CONGRUENCE 
Tests  for  congruent  triangles : 

Any  two  triangles  are  congruent  if: 

1.  Two  sides  and  the  included  anglj §3o,  Th.  1 

2.  Two  angles  and  the  included  side §36,  Th.  2 

3.  Three  sides §39,  Th.  4 

301 


302  PLANE  GEOMETRY 

Two  right  triangles  are  congruent  if: 

4.  The  hypotenuse  and  an  acute  angle §81,  Th.  22 

5.  The  hypotenuse  and  a  side §82,  Th.  23 

Test  for  congruent  parallelograms: 

Two  parallelograms  are  congruent  if  two  sides  and  the 

included  angle §98,  Th.  35 

General  test  for  congruent  figures : 

Any  two  figures  are  congruent  if  they  can  be  made  to  coincide  ....  §33 

Tests  for  equal  angles : 

Two  angles  are  equal  if  they  are: 

1.  Sums,  differences,  equal  multiples,  or  equal  parts  of 

equal  angles §  §30,  47 

2.  Right  angles  or  straight  angles §§22,  29 

3.  Supplements  or  complements  of  equal  angles §§24,  25,  29 

4.  Vertical  angles §§28,  29 

5.  Corresponding  angles  of  congruent  triangles §38 

6.  Base  angles  of  an  isosceles  triangle §37,  Th.  3 

7.  Alt.  int.  angles  of  parallel  Hnes §68,  Th.  14 

8.  Corresponding  angles  of  parallel  lines. §69,  Th.  15 

9.  The  third  angles  of  two  triangles  that  have  two 

angles  of  one  equal  respectively §75,  Cor.  II 

10.  Angles  with  their  sides  parallel  right  side  to  right  side .  §73,  Ex.  3 

11.  Angles  with  their  sides  perpendicular §86,  Exs.  38,  39 

12.  Opposite  angles  of  a  parallelogram §95,  Th.  32 

13.  Ceatral  angles  subtended  by  equal  arcs §135,  As.  50 

14.  Central  angles  subtended  by  equal  chords §136,  Th.  61 

15.  Angles  measured  by  equal  arcs §173,  II 

16.  Corresponding  angles  of  similar  figures §§210,  256 

17.  Angles  of  a  regular  polygon §§78,  270 

Construction  of  equal  angles §40,  Prob.  1 ;  §41,  Prob.  2 

Tests  for  equal  segments: 

Two  segments  are  equal  if  they  are: 

1.  Sums,  differences,  equal  multiples,  or  equal  parts  of 

equal  segments §§30,  47 

2.  Radii  of  the  same  or  equal  circles §§12,  29 

3.  Sides  of  an  isosceles  triangle §34 

4.  Corresponding  sides  of  congruent  triangles §38 

5.  Opposite  sides  of  a  parallelogram §95,  Th.  31 

6.  Parallel  or  perpendicular  segments  between  par- 
allels  §104,  Ths.  39,  40 


OUTLINE   SUMMARY  303 

Equal  segments  are  formed  when: 

7.  The  diagonals  of  a  parallelogram  intersect §95,  Th.  33 

8.  A  series  of  parallels  cutting  equal  segments  on  one 
transversal  intersect  a  second  transversal §111,  Th.  45 

9.  Perpendiculars  are   drawn   from   a  point  in   the 
bisector  of  an  angle  to  the  sides  of  the  angle.  .  .  .  §179,  Th.  85 

10.  A  point  in  the  perpendicular  bisector  of  a  segment 

is  joined  to  the  extremities  of  the  segment §180,  Th.  86 

11.  A  radius  is  perpendicular  to  a  chord §137,  Th.  63 

12.  Perpendiculars  are  drawn  from   the  center  of  a. 

circle  to  two  equal  chords §141,  Th.  67 

13.  Two  tangents  are  drawn  to  a  circle  from  a  point 
without §146,  Th.  70 

14.  Three  terms  of  one  proportion  are  equal  respec- 
tively to  three  terms  of  another  proportion.  ..  §201,  Th.  92 

Division  of  a  segment  into  equal  parts §111,  Prob.  7 

Tests  for  equal  arcs: 

Two  arcs  are  equal  if: 

1.  They  have  equal  central  angles §135,  As.  49 

2.  They  have  equal  chords §136,  Th.  62 

3.  A  radius  is  perpendicular  to' the  chord  of  an  arc  . .  §137,  Th.  63 

4.  They  are  intercepted  by  parallel  chords §166,  Th.  80 

5.  They  are  intercepted  by  a  chord  and  a  tangent 

parallel  to  it §171,  Th.  82 

6.  They  measure  equal  angles §173,  II 

Tests  for  equal  chords : 

Two  chords  are  equal  if: 

1.  They  have  equal  central  angles §136,  Th.  61 

2.  They  have  equal  arcs §136,  Th.  62 

3.  They  are  equally  distant  from  the  center §140,  Th.  66 


SIMILARITY 

Tests  for  similar  triangles : 

Two  triangles  are  similar  if: 

1.  They  are  mutually  equiangular §210,  Th.  102;  §257 

2.  An  angle  of  one  equals  an  angle  of  the  other  and 

the  sides  including  the  angle  are  proportional ....  §258,  Th.  119 

3.  Corresponding  sides  are  proportional §259,  Th.  120 


304  PLANE  GEOMETRY 

Tests  for  similar  polygons : 
Two  polygons  are  similar  if: 

1.  The  angles  of  one  are  equal  respectively  to  the 
angles  of  the  other  and  the  corresponding  sides  are 
proportional §256 

2.  Diagonals  from  corresponding  vertices  divide  the 
polygons  into  triangles  that  are  similar  and  simi- 
larly placed §261,  Th.  121 

3.  They  are  regular  polygons  of  the  same  number  of 

sides §287,  Th.  134 

Properties  of  similar  figures: 

1.  The  corresponding  angles  are  equal. §§210,  256 

2.  Corresponding  sides  have  equal  ratios §§210,  256 

3.  Diagonals  from  corresponding  vertices  divide  the 
polygons  into  triangles  that  are  similar  and  simi- 
larly placed §263,  Th.  122 

4.  The  ratio  of  corresponding  segments  equals  the 

ratio  of  simiHtude §264,  Ths.  123,  125 

5.  The  ratio  of  the  areas  equals  the  square  of  the 

ratio  of  similitude §266,  Th.  126;  §267,  Th.  127 

Equal  ratios  and  circles: 

1.  The  ratio  of  the  circumferences  of  any  two  circles 

equals  the  ratio  of  the  diameters  or  of  the  radii .  .  §307,  Th.  140 

2.  The  ratio  of  the  areas  of  any  two  circles  equals  the 
ratio  of  the  squares  of  the  diameters  or  of  the  radii 

§307,  Th.  141 

Tests  for  equal  ratios  or  equal  products §211 

Two  ratios  or  two  products  are  equal  when : 

1.  Parallels  cut  two  transversals 

§203,  Th.  98;  §204,  Th.  99  and  Cor. 

2.  Polygons  are  similar §210 

3.  Two  ratios  are  equal  to  a  third  ratio §199,  As.  57 

Construction  of  proportional  segments: 

1.  The  division  of  a  segment  into  parts  proportional 

to  any  number  of  given  segments §206,  Prob.  15 

2.  The  fourth  proportional  to  three  given  segments  §207,  Prob.  16 

3.  The  mean  proportional  to  two  given  segments  .§221,  Prob.  17 

Important  cases  of  equal  ratios  occur  when: 

1.  Two  chords  intersect  within  a  circle §213,  Th.  103 

2.  Two  secants  intersect  without  a  circle. §215,  Th.  104 


OUTLINE  SUMMARY  305 

3.  A  secant  and  a  tangent  intersect  without  a  circle.  §216,  Th.  105 

4.  A  line  bisects  an  angle  of  a  triangle §217,  Th.  106 

5.  A  line  bisects  an  exterior  angle  of  a  triangle §218,  Th.  107 

6.  A  perpendicular  is  drawn  from  the  vertex  of  the 
right  angle  of  a  right  triangle  to  the  hypotenuse 

§220,  Ths.  108,  109 

EQUIVALENCE 

Tests  for  equivalence §246,  Ths.  IIG,  117 

Two  parallelograms  or  two  triangles  are  equivalent  or  a 
triangle  is  half  of  a  parallelogram  if: 

1.  They  have  the  same  base  and  the  same  altitude. 

2.  The  product  of  the  base  and  altitude  of  one,  etc. 
Any  two  polygons  are  equivalent  if  they  are: 

Sums,  differences,  or  equal  parts  of  equivalent  figures. 

Construction  of  equivalent  figures: 

1.  To  transform  a  parallelogram  into  a  rectangle  on  a 

given  base §248,  Prob.  18 

2.  To  transform  a  parallelogram  into  a  square §248,  Prob.  19 

3.  To  transform  a  polygon  into  a  triangle.  . .  .  §249,  Probs.  20,  21 

4.  To  construct  a  square  equal  to  the  sum  of  two 

squares §251,  Prob.  22 

5.  To  construct  a  square  equal  to  the  difference 
between  two  squares §251,  Prob.  23 

MEASUREMENT 
Meastirement  of  angles: 

1.  Of  central  angles §157,  As.  53 

2.  Of  inscribed  angles §161,  Th.  77 

3.  Of  an  angle  formed  by  a  chord  and  a  tangent.  .  §170,  Th.  81 

4.  Of  an  angle  formed  by  two  chords  that  intersect.  §164,  Th.  78 

5.  Of  an  angle  formed  by  two  secants,  two  tangents, 

or  a  secant  and  a  tangent  ..§165,  Th.  79;  §172,  Ths.  83,  84 

6.  By  trigonometric  ratios §225 

Angle-sums: 

The  sum  of 

1.  Adj.  angles  on  one  side  of  a  st.  line  having  a 
common  vertex  is  2  rt.  -4 §§26,  29 

2.  Adj.  angles  about  a  point  is  4  rt.  A §§26,  29 

3.  Int.  angles  on  one  side  of  a  transversal  when  two 
parallels  are  cut  b}'  a  third  st.  line  is  2  rt.  zi §69,  Th.  16 

4.  The  interior  angles  of  a  triangle  is  2  rt.  ^ §74,  Th.  18 


306  PLANE   GEOMETRY 

The  sum  of 

5.  Two  angles  of  a  triangle  is  equal  to  the  opposite 

exterior  angle §76,  Th.  19 

6.  The  acute  angles  of  a  right  triangle  is  1  rt.  Z . .  .  .  §75,  Cor.  Ill 

7.  The  interior  angles  of   a  polygon  of  n  sides  is 

2{n—2)  rt.  A §79,  Th.  20 

8.  The  exterior  angles  of  any  polygon  is  4  rt.  A §80,  Th.  21 

Measurement  of  polygons: 

1.  The  area  of  a  rectangle  is  ab §239,  As.  63 

2.  The  area  of  a  parallelogram  is  ah §242,  Th.  113 

3.  The  area  of  a  triangle  is  ^  a& §243,  Th.  114 

"  y2  he  sin  ^ .  . .  .^.  .  .  ^.  . .  §253,  Ex.  41 
"  ^|s{s-a){s-  h)  {s'-c) . .  §253,  Ex.  44 

4.  The  area  of  a  trapezoid  is  3^o(6+6') §244,  Th.  115 

5.  The  area  of  a  regular  polygon  is  %  per.   X 

apothem §286,  Th.  133 

6.  For  the  area  of  irregular  polygons  see §245 

Measurement  of  circles  and  sectors : 

1.  The  circumference  of  a  circle  is  2itr §301,  Th.  137 

2.  The  area  of  a  circle  is  tt^-^ §304,  Th.  138 

3.  The  area  of  a  sector  is  -rr-TTr §305,  As.  66 

3b0 

ELEMENTARY  FIGURES 
Properties  of  triangles: 

1.  The  sum  of  the  angles  of  a  triangle  is  2  rt.  A §74,  Th.  18 

2.  Thd  angle  opposite  the  greater  side  of  a  triangle.  §129,  Th.  55 

3.  The  side  opposite  the  greater  angle  of  a  triangle.  .  §128,  Th.  54 

4.  The  medians  are  concurrent §115,  Th.  49 

5.  The  perpendicular  bisectors  of  the  sides  are  con- 
current  §184,  Th.87 

0.  The  altitudes  are  concurrent §184,  Th.  88 

7.  The  bisectors  of  the  angles  are  concurrent §185,  Th.  89 

Construction  of  triangles: 

Two  sides  and  an  angle  opposite  one §55,  Ex.  4 

Properties  of  isosceles  triangles: 

1.  Two  sides  are  equal §34 

2.  The  base  angles  are  equal §37,  Th.  3 

3.  Bisector  of   vertex  angle,    the  altitude,   and   the 
median  to  the  base  coincide 

§50,  Th.  6;  §51,  Th.  7;  §84,  Th.  25 


OUTLINE  SUMMARY  307 

Tests  for  isosceles  triangles : 

A  triangle  is  isosceles  if : 

1.  Two  sides  are  equal §§34,  83 

2.  Two  angles  are  equal §83,  Th.  24 

Properties  of  right  triangles: 

1.  The  acute  angles  are  cornplcnienLary §75,  Cor.  Ill 

2.  The  median  from  the  vertex  of  the  right  angle  is 
one-half  the  hypotenuse §116,  Th.  50 

3.  If  a  and  b  are  the  legs  and  c  is  the  hypotenuse, 

a2+62=c2 §222,  Th.  110;  §251,  Th.  118 

Properties  of  parallelograms: 

1.  The  opposite  sides  are  parallel §95 

2.  The  opposite  sides  are  equal §95,  Th.  31 

3.  The  opposite  angles  are  equal §95,  Th.  32 

4.  The  diagonal  bisects  the  parallelogram §95,  Th.  30 

5.  The  diagonals  bisect  each  other §96,  Th.  33 

Properties  of  special  parallelograms §§105-110 

Tests  for  parallelograms: 

A  quadrilateral  is  a  parallelogram  if: 

1.  Each  side  is  parallel  to  its  opposite §99 

2.  One  side  is  equal  and  parallel  to  its  opposite §100,  Th.  36 

3.  Each  side  is  equal  to  its  opposite '.  .  §101,  Th.  37 

4.  The  diagonals  bisect  each  other §102,  Th.  38 

Properties  of  regular  polygons: 

1.  The  sides  and  angles  are  equal §§78,  270 

2.  A  circle  can  be  circumscribed  about  the  polygon . .  §282,  Th.  130 

3.  A  circle  can  be  inscribed  in  the  polygon §283,  Th.  131 

4.  The  radius  bisects  the  angle §282,  Th.  130  Cor. 

2»7  —4 

5.  Each  angle  is  rt.  A §285,  Th.  132 

n 

Tests  for  regular  polygons : 

A  polygon  is  regular  if : 

1.  The  sides  and  angles  are  equal §§78,   270 

2.  A  circle  is  divided  into  equal  arcs  and 

a.  The  points  of  division  are  joined §271,  Th.  128 

b.  Tangents  are  drawn  to  the  points  of  division  §271,  Th.  129 

Construction  of  regular  polygons §§273-280,  290 


308  PLANE  GEOMETRY 

INEQUALITIES 

Tests  for  unequal  segments : 

1.  The  sum  of  two  sides  of  a  triangle,  etc §126,  Ass.  37,  38 

2.  If  one  angle  of  a  triangle  is  greater  than  another, 

etc §128,  Th.  54 

3.  The  perpendicular  is  the  shortest  distance,  etc. .  .  .  §130,  Th.  56 

4.  If  from  a  point  in  a  perpendicular  to  a  line  oblique 
segments  are  drawn  cutting  off  unequal  distances, 

etc §130,  Th.  57 

5.  If  from  a  point  in  a  perpendicular  to  a  line  two 
unequal  oblique  segments  are  drawn,  etc §130,  Th.  58 

6.  If  two  triangles  have  two  sides  of  one  equal  to  two 

sides  of  the  other  but  the  included  angles,  etc. .  .  §131,  Th.  59 

7.  If  from  a  point  within  a  triangle  segments  are 

drawn  to  the  extremities  of  one  side §127,  Th.    .53 

Tests  for  unequal  angles: 

1.  The  exterior  angle  of  a  triangle,  etc §125;  §58,  Th.  8 

2.  If  one  side  of  a  triangle  is  greater  than  another, 

etc §129,  Th.  55 

3.  If  two  triangles  have  two  sides  of  one  equal  to  two 

sides  of  the  other  but  the  third  side  of  one,  etc. .  .  §132,  Th.  60 


INDEX 


[References  are  to  page  numbers) 


Abbreviations 19 

Acute  angle 9 

Addition  or  composition.  .  .  .    167 

Addition:  of  angles 7 

of  polygons 209 

Adjacent  angles 7 

Algebraic  analysis 222 

Algebraic  equations  indicating 

constructions  197;  227,  Ex.  5 
Algebraic  notation  in  proof s .      51 

Ahmcs 283 

Alternate  exterior  angles ....     49 
Alternate  interior  angles.  ...     49 

Alternation:  extreme 166 

mean 166 

Altitude:  length  of,  in  equi- 
lateral triangle 190 

of  parallelogram 85 

of  trapezoid 84 

of  triangle 70 

Altitudes  of  triangle,  concur- 
rent     150 

ratio  of 244 

Analysis,  to  make  an 25 

Angle 6 

acute 9 

arms  of 6 

bisector  of 8,  31 

central,  in  circle 108,  124 

central,    of    regular    poly- 
gon  260,  261 

complement  of 12 

degree  of 11 

designation  of 6 

division  of 31 

exterior,  of  triangle 48,  61 

21  309 


formed  by  rotation.. 6 

included 22 

inscribed  in  an  arc 128 

inscribed  in  a  circle 126 

measurement  of 11,  137 

obtuse 9 

of  elevation 196 

of  sixty  degrees 60 

of  regular  polygon 262 

re-entrant 63 

right 9 

right  and  left  sides  of 58 

sides  of 6 

size  of 6 

straight 9,  16 

supplement  of 12 

trisection  of 32 

vertex  of 6 

vertex,  of  isosceles  triangle.  20 

Angles:  addition  of 7 

adjacent 7 

alternate  exterior 49 

alternate  interior 49 

complementary 12 

congruent 7 

consecutive    of    parallelo- 
gram    79 

construction  of  equal 7,  30 

corresponding,   of   congru- 
ent figures 20 

corresponding,  of  lines  cut 

by  a  transversal 49 

equal,    definition    of    (See 

Equal  angles) 11 

exterior,  of  lines  cut  by  a 

transversal 49 


310 


INDEX 


Angles  {continued) : 

interior  non-adjacent 48 

interior,  of  lines  cut  by  a 

transversal 49 

made     by     parallels     and 

transversals 54 

opposite 15 

opposite,     of     parallelo- 
gram   79»  80 

subtraction  of 8 

sum  of,  in  polygon  .......     64 

sum  of,  in  triangle 59 

supplementary 12 

symmetric 75 

vertical 15 

Angles  and  parallels 54,  57 

Angle-sums 16,  305 

Antecedent 163 

Apothem 260 

Apothems,  ratio  of 263 

Approximate  constructions.  .  282 
Approximate    measure :    of 

length 161,  162 

of  surface  of  rectangle  .212,  214 

of  heights  and  distances. .  .    196 

Approximate  value  of  7r..272,  283 

Arab 98 

Arc 5 

degree  of 123 

intercepted,      of       central 

angle 108 

intercepted,     of     inscribed 

angle .    126 

major 108 

measurement  of.  ..108,  123,  137 

minor 108 

Archimedes 220,  283 

Architecture,  exercises  from 
gable  {See  Carpentry ; 
Church  windows;  Gothic 
arch ;  Mouldings ;  Rafters ; 
Roof  trusses;  Steel 
beams) 


Arcs:  congruent 108 

equal  {See  Equal  arcs)  ....    108 

Area 208 

assumptions         concerning 

211,  275,  290 
maximum 284,  290 

Area:  of  circle 275,  277 

of  irregular  polygons 219 

of  kite 229,  Ex.16 

of  parallelogram 216 

of  rectangle 211 

of  regular  polygon 262 

of  rhombus 228,  Ex.  5 

of  sector  of  circle 275 

of  segment  of  circle 276 

of  trapezoid 218 

of  triangle 217,  228 

Areas,   ratio  of 

237,  Ex.  34;  246,  247,  263, 276 

Assumption 35 

Assumptions   concerning; 

angles 16,  17 

angle-sums i6 

area 211,  275,  290 

circles 16,  109,  118, 

124,  125,  274, 
275,  290,  291 

congruence 20 

equal  angles 16 

equivalence 210 

inequality 48,  99,  100 

location  of  lines,  rays,  and 

segments 15 

location  of  points 15 

maximum  area 290 

minimum  perimeter 291 

parallels 53 

perimeters 274,  291 

ratios 165 

sectors 275 

segments 16,  17 

straight  angles 16 


INDEX 


311 


Axial  symmetry 76 

relation    to    central    sym- 
metry       78 

Axis  of  symmetry 76 

of  circles 1 18,  1 19 

Base  of  an  isosceles  triangle .     20 

Bases :  of  parallelogram 79 

of  trapezoid 84 

Bisection:  of  an  angle 8,  31 

of  a  polygon 209 

of  a  segment 4,  35 

Bisector  of  an  angle:  as  locus  146 

construction  of 8,  31 

segments  made  by .  .  .  .  185,  186 

Bisectors:     of    angles    of    a 

triangle,  concurrent ...  .    150 
perpendicular  (See  Perpen- 
dicular bisector) 

Carpentry,     exercises     from 

31,  129,  157,  I75»  265,  266 

Center:  of  a  circle 5 

of  gravity 155 

of  regular  polygon 260 

of  similitude 249 

of  symmetry 77 

of  symmetry  of   parallelo- 
gram      81 

Centers,  line  of  {See  Line  of 

centers) 117 

Centers  of  circles,  loci  of .  .  .    158 
Central  angle :  of  a  circle  {See 
equal    angles;     Unequal 

angles) 108 

measure  of 124 

Central  angle  of  a  regular 

polygon 260 

measure  of 261 

Central  symmetry 77 

relation  to  axial  symmetry     78 

test  for 77 

Chinese 189,  283 


Chord 5 

common      {See     Common 

chord) 117 

fundamental  theorem  of  .  .    1 1 1 
Chord  and  tangent,  measure 

of  angle  made  by 135 

Chords:  equal,  tests  for  {See 

Equal  chords) 303 

intersecting  {See  Intersect- 
ing chords) 

parallel 132 

Church  windows 

140,  158,  201,  280 

Circle 5 

arc  of 5 

area  of 275,  277 

central  angle  of 108 

chord  of 5 

circumference  of 269,  274 

circumscribed 120,  151 

circumscribed,    of    regular 

polygon 259 

construction  of.  .    151,  154,  156 

definite  location  of 113 

diameter  of. 5 

escribed 151 

inscribed. 120,  151 

inscribed  angle  of 126 

inscribed,  of  regular  poly- 
gon    260 

measurement  of 277,  306 

radius  of 5 

sector  of 157,  275 

segment  of 276 

tangent  to 114,  134 

Circles:  assumptions  concern- 
ing  16,  109,  118, 

124,  125,  274» 
275,  290,  291 

concentric 117 

congruent 108 

construction  of.  .  .J 51,  154,  156 
inequalities  in 125 


312 


inde: 


Circles  {continued) : 

intersecting Ii8 

loci  of  centers  of 153 

tangent 117,  119 

tangents  to 137 

Circles  and  equal  ratios 276 

Circles  and  symmetry 

109,  118,  119 

Circular  segments 276 

Circumference 269,  274 

formula  for 274 

ratio  of,  to  diameter 276 

Circumferences,  ratio  of 276 

Circumscribed  circle 120 

construction  of 151 

of  regular  polygons 259 

Circumscribed  polygon 120 

Coincident  rays 5 

Collinear  rays 5 

Commensurable  segments.  .  .  164 

Common  chord 117 

as  bisector '. .  .  119 

bisected 118 

Compasses:  proportional 191 

use  of 6,  32 

Complement 12 

Complementary  angles 12 

Composition  or  addition.  ...  167 

Compound  curves 157 

Computations:   of  areas 

214,  228,  277 
by  trigonometric  ratios 

194,  196,  202 
measurement     of     regular 

polygons 267 

of    perimeters    of    regular 
inscribed  polygons.  .270,  272 

Concave  polygon 63 

Concentric  circles 117 

Conclusion 36 

Concrete  representation:    of 

points I 

of  straight  lines I 


Coiicurrcnt  lines 91 

exercises  involving 155 

special  cases  of .  .  .  115,  148,  150 

Congruence 20,  301 

Congruent  angles 7 

arcs 108 

circles 6,  108 

figures 20 

parallelograms 81 

segments 4 

triangles 301 

Consecutive  angles 79 

Consecutive  sides 79 

Consequent 163 

Construction  lines 57 

Construction:    of     equal 

angles 7,  30 

of  bisector  of  an  angle.  .  .  .8,  31 

of  circles 151,  154,  156 

of  circumscribed  circle ... .    151 

of  decagon,  regular 256 

of  equal  segments 88 

of  equivalent  figures 

210,  222-224,  227 

of  escribed  circle 151 

of  extreme  and  mean  ratio.  256 

of  fourth  proportional 1 73 

of  hexagon,  regular 254 

of  inscribed  circle 151 

of  mean  proportional 187 

of  octagon,  regular 253 

of  parallels 53 

of  pentadecagon,  regular. . .   258 

of  pentagon,  regular 258 

of  perpendiculars 

9-10,  33-35,  133 
of  proportional  segments .  .  1 72 
of  regular  polygons. . .  .252-258 

of  similar  polygons 243 

of  square 253 

of  tangents 115,  134,  137 

of  triangles 

2 1 ,  Ex.  4 ;  24,  Ex. ;  47,  Ex.  4 


INDEX 


313 


Constructions:   by   algebraic 

analysis 222 

indicated  by  equations 

197;  227,  Ex.  5 

Contact,  point  of 114 

Converse  theorems 56 

Convex  polygon 63 

Corresponding  angles:  of  con- 
gruent figures 20 

of  lines  cut  by  a  transversal     49 
Corresponding  sides:  of  con- 
gruent figures 20 

of  similar  figures 176,  240 

of  similar  triangles 178 

Corollary 36 

Cosine  of  an  acute  angle.  ...  193 
Cross-sections  of  columns.  .  .  214 
Cut-glass  designs 141,  268 

Decagon    {See  Regular  deca- 
gon)      63 

Definite  location :  of  circles . .    112 

of  lines 2,  15 

of  points 2,  15 

of  rays 5,  15 

of  segments 4,  15 

Degree :  of  angle 11 

of  arc 123 

Design,  theory  and  practice 

of 71,  79,  157,  231,  256 

Determination  of  points 152 

Diagonals:  of  parallelograms 

79,  80 

of  polygons 63 

of  quadrilaterals 80,  81 

of  square,  formula  for 190 

Diagrams  for  review 43 

Diameter  of  circle 5 

tests  for 1 12,  1 16 

Diameters,  ratio  of 276 

Difference     :    between    two 

angles 8 

between  two  polygons ....   209 


Direct  proof 51 

Distance :  between  two  points      4 

from  a  point  to  a  line 104 

Division  or  subtraction 167 

Division  of  angles 31 

Division  of  segments:  exter- 
nal and  internal 182 

harmonic 186 

in  extreme  and  mean  ratio 

255,  256 

in  equal  parts 88 

in    parts    proportional    to 

given  segments 172 

Draftsman's  methods,  me- 
chanical drawing 

9,  10,  32,  53,  259,  282 
Duodecagon      {See     Regular 

duodecagon) 63 

Egypt,  Egyptians 

I,  69,  189,  191,  283 

Elements,  Euclid's 45 

Elevation,  angle  of 196 

Engineering,  problems  from 
{See  Architecture ;  Sur- 
veying ;      Railroading) 

279,  291 

Equal  angles 1 1 

assumptions  concerning.  .  .      16 

tests  for 16,  37,  302 

Equal  arcs 108 

have  equal  central  angles. .    109 

have  equal  chords no 

measure  equal  angles 137 

measured  by  equal  angles..  137 

tests  for 303 

Equal    central    angles    of    a 

circle:  have  equal  arcs.  .    109 

have  equal  chords no 

Equal  chords:  equally  distant 

from  center 113 

have  equal  arcs no 

have  equal  central  angles. .   no 


314 


INDEX 


Equal   products:   exercises 

involving 179,  197 

important  cases  of 182 

tests  for 177 

Equal  ratios:  applications  of.   191 
exercises  involving 

174,  179,  197 
important  special  cases  of  182 

series  of 245 

tests  for 177,  304 

Equal  segments 4 

•    tests  for , 302 

Equations 296 

Equilateral  triangle 20 

altitude  of,  formula  for... .    190 

angles  equal 25 

side  of,  formula  for 190 

value  of  each  angle 60 

Equivalent  polygons 209 

assumptions  concerning.  . .   210 
construction  of 

210,  220,  236,  Ex.  32;  237, 

Ex.  33;  251,  Ex.  22;  305 

exercises  involving ....  224,  233 

tests  for 210,  220,  305 

Escribed  circle 151 

Euclid 45,  137,258 

Exact  measure 161 

Exterior  angle  of  triangle 48 

Exterior  angles  of  lines  cut 

by  transversal 49 

External  division 182 

Extreme  alternation 166 

Extreme  and  mean  ratio.  .  . .   256 
Extremes 165 

Fixed  line,  ray,  or  segment. .     58 

Floor  designs 69,  79,  97, 

157,  204,  230,  235,  266,  281 
Formula:     for     altitude     of 

equilateral  triangle 190 

for  angle-sums 305 

for  diagonal  of  square 190 


for  measurement  of  circles 

and  sectors 277,  306 

for  measurement  of  poly- 
gons   228,  306 

for  side  of  equilateral  tri- 
angle     190 

for  side  of  square 190 

Fourth  proportional 173 

construction  of 173 

Fractions 292 

Fundamental  assumption:  of 
measurement  of  poly- 
gons    211 

regarding  parallels 53 

Fundamental    characteristic: 

of  parallelograms 80 

of  ratios 165 

Fundamental     relation     be- 
tween arcs  and  angles. . .   124 
Fundamental   test:    for    in- 
equality     100 

for  parallelograms. ...'.,. .     82 

for  parallels 50 

f ov  special  quadrilaterals .  .     84 
Fundamental     theorems     of 

proportion 165 

Gable 206 

Gauss 258 

General  assumptions 16 

Generation    or  formation  of 

angles 6 

Geometric  forms,  occurrence 

of 69,71,79,  106, 

158,  231,  236,  254, 
265,  266,  281,  291 
Geometrical     problem     {See 

Construction) 36 

Golden  section 256 

Gothic  arch 206,  280 

Gravitv,  center  of I55 


INDEX 


315 


Greek  geometry 

21,  2>2,  45,  59,  189,  191,  233, 
235.  256,  258,  266,  274,  283 

Harmonic  division 186 

Heptagon 63 

Hero  of  Alexandria 233 

Hexagon  {See  Regular  hexa- 
gon)       63 

Hindus 189,283 

Historical  notes:  Ahmes 283 

Arab 98 

Archimedes 220,  283 

Chinese 189,  283 

Egyptians. .  i,  69,  189,  191,  283 

Euclid 45.  137,258 

Gauss 258 

Greeks 32,  59»  274 

Hero  of  Alexandria 233 

Hindus 189,283 

Hippocrates 279 

Moderndiscoveries258,274,283 

Pappus 235 

Pi(7r) 164,  274»  275.  283 

Pythagoras 59,  189 

Pythagoreans 189,  256,  266 

Thales 27,  191 

Trisection  of  angles 32 

Hippocrates 279 

Hypotenuse 66 

Hypothesis 36 

Incenter 155 

Included  angle 22 

Included  side 23 

Incommensurable  segments. .    164 

Indirect  proof 51 

Inequalities     {See     Unequal 

angles;   Unequal  sides); 

assumptions    concerning 

48,99 

fundamental  test  for 100 

in  circles 125 


Inscribed  angle  in  an  arc. .. .    128 
Inscribed  angle  in  a  circle. . .   1 26 

measure  of 126 

Inscribed  angle  in  a  semicircle  128 

Inscribed  circle 1 20 

construction  of 1 51 

in  regular  polygons 260 

Inscribed  polygon 120 

Inscribed    regular    polygons 

253-258 

Integraph 274 

Intercepted  arc 108,  126 

Interior  angles:  of  a  polygon     64 

of  a  regular  polygon 262 

of  lines  cut  by  a  transversal    49 

Internal  division 182 

Intersecting  circles 118,  119 

Intersecting  chords:  measure 

of  angle 130 

of  product  of  segments  of. .  1 82 
Intersecting  loci,  use  of:  in 

determination  of  points .    152 
in  construction  of  circles.  .    1 54 
Intersecting  secant  and  tan- 
gent:  measure  of  angles  of  136 
product  of  segments  of . .  .  .    184 
Intersecting  secants:  measure 

of  angle  of 131 

product  of  segments  of . .  .  .   183 
Intersecting     tangents:     are 

equal 1 1 5 

measure  of  angle  of 13^ 

Inverse  proportion 204 

Inversion 166 

Irrational  numbers 164,  274 

Irregular  polygons,  area  of.. .  219 

Isosceles  trapezoid 84 

legs  of 84 

properties  of 86 

Isosceles  triangle 20 

base  angles  of 24 

properties  of 3^6 

tests  for ..68,307 


31G 


INDEX 


Kites. 84 

area  of 229,  Ex.  16 

properties  of 86 

Legs :  of  an  isosceles  trapezoid     84 

of  a  right  triangle 66 

Length :  of  circle 269 

of  segment 4,  161 

units  of 300 

Leveling  device 41 

Limiting  values  of  perimeter 
of  inscribed  and  circum- 
scribed regular  polygons  274 
Limits,     exercises    involving 

40,  57,  67,  103,  134,  136 
Line:    parallel    to    base    of 

triangle §9,  90,  170 

straight  {See  Straight  line) 

Line  of  centers 117 

as  axis  of  symmetry 118 

as  bisector 118 

Lines :    concrete    representa- 
tion of I 

concurrent     {See    Concur- 
rent lines) 91 

construction 57 

definite  location  of 15 

parallel  {See  parallel  lines) .     50 
perpendicular  {See  perpen- 
dicular lines) 9 

symmetric 75 

Location,  definite  {See  Defi- 
nite location) 

Loci 143 

complete  proofs  for 145 

finding  of 143 

intersecting 152,  154 

miscellaneous  exercises  on .    1 59 

of  centers  of  circles 1 53 

of  points 146 

of  vertices  of  triangles 

159,  Ex.  4 


Major  and  minor  arc 108 

Maximum  and  minimum. ..  .   284 

Mean  alternation 166 

Mean  proportional 179 

construction  of 187 

Mean  ratio,  extreme  and. ...  256 

Means 165 

Measure  (measurement):  ap- 
proximate    161 

exact 161 

of  angles ...11,  137,  305 

of  arcs .■ 123,  137 

of  circles  {See  Circles) 

269,  277,  306 
of  polygons  {See  Polygons) 

211-220,  306 

of  segments 161 

of  surfaces 208 

practical 162,  214 

Measure  number :  of  angles .  .     11 

of  a  segment 161 

of  a  surface 208 

Mechanical  drawing 

9,  10,  32,  53,  259,  282 
Median:  of  right  triangle. ...     92 

of  a  triangle 42 

of  a  quadrilateral 85 

Medians  of  a   triangle  con- 
current       91 

Mid-point  of  .segment,  deter- 
mination of 4,  35 

Minimum 284 

Minor  arc 108 

Minutes n,  123 

Modern  discoveries  in  geom- 
etry  258,  274,283 

Moldings I57 

Nature     of     theorems     and 

proofs 35 

Navigation,  exercise  from... .     71 

Numbers:  irrational 164,  274 

ratio  of 163 


INDEX 


317 


Obtuse  angle 9 

Octagon    {See  Regular  octa- 
gon)       63 

Opposite  angles 15 

Opposite  angles  of  a  parallelo- 
gram  79.  80 

Opposite  sides  of  a  parallelo- 
gram   79.  80 

Origin  of  ray 5 

Orthocenter i55 

Pappus 235 

Parallel  chord  and  tangent ...    136 

Parallel  chords 132 

Parallel  lines 50 

construction  of 53 

fundamental    assumption 

regarding 53 

tests  for .* 53.  301 

Parallel  rulers 95 

Parallel  to  base  of  a  triangle 

89,  90,  170 
Parallels    and     transversals : 

angles  formed  by 54.  55 

equal  segments  formed  by 

88,  98 
proportional  segments 

formed  by 168 

Parallelogram 79.  85 

altitude  of 85 

area  of 216 

bases  of 79.  85 

consecutive      angles     and 

sides  of 79 

diagonals  of 63,  79,  80,  81 

opposite  angles  and  sides  of     79 
Parallelograms :    congruence 

of 81 

properties  of 307 

test  for 82,  83,  307 

Pencil  of  rays 5 

Pentadecagon    (See    Regular 

pentadecagon) 63 


Pentagon  (See  Regular  penta- 
gon)       63 

Pentagram  star 266 

Perigon 9 

Perimeter 63 

minimum 285,  289,  291 

ratio  of,  to  diameter 263 

Perimeters:  computation  for 

270,  271 

lengths  of 272 

ratio  of 245,  263 

Perpendicular     bisector :     as 

locus 147 

construction  of 10.  35 

Perpendicular  bisectors  con- 
current     149 

Perpendicular  lines 9 

construction  of  .9,  10,33-35,  ^33 

tests  for 301 

Physics,  exercises  from.  .  ig6,  138 

Pi  (tt) 273 

approximate  constructions 

for 282 

approximate  value  of.  .273,  283 

construction  of 274 

history  of 283 

is  irrational 164,  274 

Point:  determination  of  mid- 
point   4,  35 

of    contact    of    line    and 

circle 114 

of     contact      of     tangent 

circles 1 20 

of  tangency 114 

variable 143 

Points:   concrete  representa- 
tion of I 

definite  location  of 2,  15 

determination  of 1 52 

locus  of 146 

symmetric 75,  118 

Polygon 63 

area  of 208 


318 


INDEX 


Polygon  (continued) : 

bisection  of 209 

circumscribed 1 20 

concave 63 

convex 63 

diagonal  of 63 

inscribed 1 20 

perimeter  of 63 

regular  (See  Regular  poly- 
gon)      64 

sides  of 63 

star  (See  Star  polyrons)  141,  268 

surn  of  angles  of 64 

trisection  of 209 

vertices  of 63 

Polygons:  addition  of 209 

congruence  of  (See  Con- 
gruent figures;  Con- 
gruent parallelograms; 
Congruent  triangles) ....     20 

difference  between 209 

equivalence  of  {See  Equiva- 
lence)   220,  305 

measurement  of 

211-219,  228,  306 

names  of 63 

regular  (See  Regular  poly- 
gon)   , 252 

similar  (See  Similar  figures)  240 
star  (See  Star  polygons) 

141,  268 

subtraction  of 209 

sum  of 209 

transformation  of 210,  221 

Pons  asinorum 46 

Practical  measurements.  .162,  214 
Problem,     geometrical     (See 

Construction) 36 

Products,  test  for  equal  (See 

Equal  products) 178 

Projection 239 

Proof 36 

by  superposition 36 


direct  synthetic 36,  51 

indirect 51 

Properties:  of  isosceles  trape- 
zoids       86 

of  isosceles  triangles 306 

of  kites 86 

of  parallelograms 307 

of  rectangles 86 

of  regular  polygons 307 

of  rhombuses 87 

of  right  triangles 307 

of  squares 87 

of  similar  polygons 304 

of  trapezoids 93 

of  triangles • 306 

Proportion 165 

by  addition 167 

by  alternation 166 

by  composition 167 

by  division 167 

by  inversion 166 

by  subtraction 167 

fundamental  theorems  of . .    165 

inverse 204 

reciprocal 204 

Proportional  compasses  (See 
Compasses) 

Proportional:  fourth 173 

mean 179,  187 

Proportional    segments    (See 

Equal  ratios) 178 

Construction  of 172,  304 

special  cases  of 182,  304 

Proportionally,  divided.'.  ...    178 

Proportions,  transformations 

of 166 

Protractor 124 

Pythagoras 59,  189 

Pythagorean  theorem 1 89 

exercises  involving 

189,  199,  238 


INDEX 


3m 


Pythagorean  theorem  (cont'd) : 

proofs  for i88;  206,  Ex.  22; 

226;  235,  Ex.  22;  238 
related  theorems 

239,  Exs.  II,  12;  251,  Ex.  21 
Pythagoreans 189,  256,  266 

Quadrilateral 63 

Quadrilaterals,  special 84 

Radially  placed 249 

Radii,  ratio  of 263,  276 

Radius:  of  a  circle 5 

of  a  regular  polygon. .  . 260,  261 
Rafter    designs,     decorated 

(See  Truss).: 157.  158 

Railroading,    exercises    from 

142,  270,  280 

Ratio:  extreme  and  mean 256 

of  altitudes 244 

of  apothems 263 

of  areas 

237,  Ex.  34;  246,  247,  263,  276 
of    circumference    to    dia- 
meter    276 

of  circumferences 276 

of  corresponding  sides.  176,  240 

of  diameters 276 

of  perimeter  to  diameter .  .    263 

of  perimeters 245,  263 

of  radii 263 

of     segments     made     by 

parallels 168,  170 

of  similitude 240 

of  two  numbers 163 

Ratios:  assumptions  concern- 
ing     165 

equal    (See    Equal   ratios) 
fundamental  characteristic 

of 165 

trigonometric 192 

Ratios  and  circles 276,  304 

Ray 5 


fixed 58 

origin  of 5 

Rays:  coincident 5 

collinear 5 

definite  location  of 15 

pencil  of 5 

Reciprocally  proportional 204 

Rectangle 85 

area  of 211 

exercises  involving  area  of  214 
properties  of 86 

Rediictio  ad  ahsurdum 51 

Re-entrant  angle 63. 

Regular    decagon :    construc- 
tion of 256 

exercises  involving 267 

Regular     duodecagon :     con- 
struction of 254 

exercise  involving 267 

Regular    hexagon,    construc- 
tion of 254 

exercises  involving. 69,  265,  267 
occurrence  of 69,  265, 

Regular    octagon :    construc- 
tion of 253 

exercises  involving  265,  266,  267 
occurrence  of 265,  266 

Regular  pentadecagon,   con- 
struction of 258 

Regular  pentagon:  construc- 
tion of 258' 

exercises  concerning .  .  .  266,  267 
occurrence  of 266 

Regular  polygon 64 

angle  of 262 

apothem  of 260 

area  of 262 

center  of 260 

central  angle  of 260 

radius  of 260 

Regular  polygons:  construc- 
tion of 253-258,  265 

occurrence  of 254 


320 


INDEX 


Regular  polygons  {continued) : 

perimeters  of 270-272 

properties  of. 259,  260,  265,  307 

measurement  of 262,  267 

similar 262 

tests  for 307 

Representation  of  points  and 

straight  lines I 

Review  diagrams 43 

Rhombus 85 

area  of 228,  Ex.  5 

properties  of 87 

Rigid  figures 41 

Right    angles    {See    Perpen- 
diculars)         9 

Right  triangle 66 

hypotenuse  of 66 

legs  of 66 

properties  of 307 

sixty-degree 250,  Ex.  9 

Roof  trusses,  exercises  based  on 
42, 127, 203, 204, 205, 234, 236 

parts  of 42,  Ex.  15 

rigidity  of 41 

Roots,  rules  for 293-295 

table  of  square 298 

Rosettes 142 

Rulers,  parallel 95 

Secant 131 

Secant  and  tangent 136,  184 

Secants,     intersecting      {See 
Intersecting  secants) 

Section,  Golden 256 

Sector  of  circles. 157 

area  of 275,  306 

Segment    joining    center    of 

circles  as  bisector 118 

bisected      {See     Line      of 

centers) 119 

Segment  of  circles 276 

area  of 276 

Segment,  straight-line 3 


bisection  of 4.  35 

definite  location  of .  ; 4,  15 

division   of    {See   Division 
of  segments) 

fixed 58 

length  of 4,  161 

measure  of 161 

mid-point  of 4»  35 

Segments:  commensurable. .  .    164 

congruent 4 

equal  {See  Equal  segments)       4 

incommensurable 164 

proportional 178,  304 

ratio  of 164 

Semicircle 108 

measure  of  angle  in 128 

Series  of  equal  ratios 245 

Sewers 279,  291 

Side  included 23 

Side:  of  equilateral  triangle. .    190 

of  square 190 

Sides:  consecutive 79 

corresponding   {See  Corre- 
sponding sides) 

of  an  angle 6 

of  a  parallelogram 79 

of  a  polygon 63 

opposite ; 79 

right  and  left,  of  an  angle .      58 
Similar   figures,    or  polygons 

176,  240 

construction  of 243 

corresponding  angles  of  176,240 
corresponding  sides  of  176, 140 
properties  of  244-247,  248,  304 
tests  for .  240-243,  248,  303,  304 
Similar  regular  polygons .  262,  264 

Similar  triangles 176 

corresponding  sides  of 178 

tests  for 240-244,  248,  303 

Similitude,  center  of 249 

ratio  of 240 


INDEX 


321 


Sine  of  an  angle 193 

Sixty-degree    right    triangle 

250,  Ex.  9 

Sixty-degrees 60 

Size  of  an  angle 6 

Special  quadrilaterals 84 

exercises  concerning 96 

Square 85 

construction  of  inscribed.  .   253 
diagonal  of,  formula  for. .  .    190 

side  of,  formula  for 190 

properties  of 87 

Square    roots     {See    Roots) 

Star  pentagram 266 

Star  polygons:  formation  of..  268 

occurrence  of 141,  265,  266 

Steel  square,  carpenter's 

31,  129,  175 

Straight  angle 9 

Straightedge 32 

Straight  line 1,2 

Straight-line    segment     {See 
Segment,  straight  line) 

Subtraction :  of  angles 8 

of  polygons 209 

Subtraction  or  division 167 

Sum  of  angles 7 

of  a  polygon 64 

of  a  triangle 59 

exterior,  of  a  polygon 65 

interior  of  parallel  lines 55 

Sum  of  polygons 209 

Sums,  angle 16,  305 

Superposition 36 

Supplement 12 

Supplementary    adjacetit 

angles 13 

Supplementary  angles 12 

Surface:  measure  of 208 

units  of 300 

Surveying,  exercises  from 

27,46,72,  157,  170,  189,191, 
196,  202 


Symbols  and  abbreviations.  .  19 

Symmetric  figures 75 

Symmetric  points 75,  118 

occurrence  of 79 

Symmetry :  axial 76 

central 77 

relation  between  axial  and 

and  central 78 

Symmetry  and  circles 

109,  118,  119 

Symmetry  and  parallelograms  8 1 

Synthetic  form 36 

Tables:  of  square  roots 298 

trigonometric 299 

Tangency,  point  of 114 

Tangent  and  chord  parallel  136 

measure  of  angle  of.» 135 

Tangent  and  secant  {See 
Intersecting  secant  and 
tangent) 

Tangent  circles 117 

tests  for 119 

Tangent  of  an  angle 193 

Tangent  to  a  circle 114 

construction  of ...  1 15,  134,  137 

tests  for 114 

Tangents  to  two  circles 138 

Tests:  for  congruence 301,  302 

for  diameters 112,  116 

for  equal  angles 302 

for  equal  arcs 303 

for  equal  chords 303 

for  equal  products. .  .  .178,  304 

for  equal  ratios 177,  304 

for  equivalence 305 

for  inequality 100 

for  isosceles  triangles 307 

for  parallels 301 

for  parallelograms 307 

for  pehpendiculars 301 

for  regular  polygons ......   307 

for  similar  polygons 304 


322 


INDEX 


Tests  (continued) : 

for  similar  triangles 303 

for  special  quadrilaterals .  .     85 

for  tangent  circles 119 

for  tangents 114 

for  unequal  angles 308 

for  unequal  segments 308 

Thales 27,  191 

Theorems 35 

converse  of 56 

proof  of 36 

Pappus' 235 

Pythagorean  (See  Pythago- 
rean theorem) 1 89 

Tiles  (See  Floor  designs) 230 

Transformation   of    polygons 

210,  221 

Transformation     of    propor- 
tions     166 

Transversals     and     parallels 
(See  Parallels) 

Trapezoid 84 

altitude  of 85 

area  of 218 

bases  of 84 

properties  of 93 

Trapezoid,      isosceles      (See 
Isosceles  trapezoid) 

Triangle 20 

altitude  of 70 

area  of 217,  228 

center  of  gravity  of 155 

centroid  of 155 

circumcenter  of 1 55 

construction  of 

21,  Ex.  4;  24;  47,  Ex.  4 


exterior  angle  of 48 

incenter  of 155 

median  of 42 

orthocenter  of 155 

properties  of 306 

sum  of  angles  of 59 

Triangle:      equilateral      (See 
Equilateral  triangle) 
isosceles  (See  Isosceles  tri- 
angle) 
right  (See  Right  triangle) 

Triangles:  congruent,  tests  for  301 

equivalent,  tests  for 305 

similar,  tests  for 303 

Trigonometric  ratios 192,  202 

Trisection  of  angles 32 

Unequal  angles,  tests  for. .  .  .  308 

Unequal  segments,  tests  for. .  308 

Units :  of  length 300 

of  measure  of  angles 1 1 

of  measure  of  arcs 108,  123 

of  surface.  . 300 

Variable  point 143 

Vertex  angle  of  isosceles  tri- 
angle   20 

Vertex  of  an  angle 6 

of  isosceles  triangle 20 

Vertical  angles 15 

of  polygons 63 

Vertices,  loci  of 159,  Ex.  4 

Width  of  board , 89 

Windo#  designs  140,  158,  201,  280 


YB  35952 


